Lanjutan InsyaAllah

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Contoh Soal Trigonometri 2

\begin{array}{ll}\\ \fbox{11}.&\textrm{Dalam}\: \: \bigtriangleup \textrm{ABC}\: \: \textrm{berlaku}\: \: \sin \displaystyle \frac{1}{2}(\textrm{A+B})=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \cos \displaystyle \frac{1}{2}\textrm{C}&&\textrm{d}.\quad -\sin \displaystyle \frac{1}{2}\textrm{C}\\\\ \textrm{b}.\quad \sin \displaystyle \frac{1}{2}\textrm{C}&\textrm{c}.\quad -\cos \displaystyle \frac{1}{2}\textrm{C}&\textrm{e}.\quad \sin \displaystyle \textrm{C} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{a}\\\\ \begin{aligned}\sin \displaystyle \frac{1}{2}(\textrm{A+B})&=\sin \displaystyle \frac{1}{2}\left ( 180^{\circ}-\textrm{C} \right )\\ &=\sin \left ( 90^{\circ}-\displaystyle \frac{1}{2}\textrm{C} \right )\\ &=\cos \displaystyle \frac{1}{2}\textrm{C} \end{aligned}.

\begin{array}{ll}\\ \fbox{12}.&\textrm{Nilai}\: \: \cos 75^{\circ}\sin 15^{\circ}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{4}\left ( 2-\sqrt{3} \right )&&\textrm{d}.\quad \displaystyle \frac{1}{4}\left ( 1-\sqrt{3} \right )\\\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\left ( 2-\sqrt{3} \right )&\textrm{c}.\quad \displaystyle \frac{1}{2}\left ( 1-\sqrt{3} \right )&\textrm{e}.\quad \displaystyle \left ( 2-\sqrt{3} \right ) \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{a}\\\\ \begin{aligned}\cos 75^{\circ}\sin 15^{\circ}&=\displaystyle \frac{1}{2}\times \left ( 2\cos 75^{\circ}\sin 15^{\circ} \right )\\ &=\displaystyle \frac{1}{2}\left [\sin \left ( 75^{\circ}+15^{\circ} \right )-\sin \left ( 75^{\circ}-15^{\circ} \right ) \right ]\\ &=\displaystyle \frac{1}{2}\left [\sin 90^{\circ}-\sin 60^{\circ} \right ]\\ &=\displaystyle \frac{1}{2}\left [ 1-\frac{1}{2}\sqrt{3} \right ]\\ &=\displaystyle \frac{1}{4}\left [ 2-\sqrt{3} \right ] \end{aligned}.

\begin{array}{ll}\\ \fbox{13}.&\textrm{Nilai}\: \: 2\cos 50^{\circ}\cos 40^{\circ}-2\sin 95^{\circ}\sin 85^{\circ}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 2\cos 10^{\circ}&&\textrm{d}.\quad 1\\\\ \textrm{b}.\quad \displaystyle 1+2\cos 10^{\circ}&\textrm{c}.\quad \displaystyle -1+2\cos 10^{\circ}&\textrm{e}.\quad -1 \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{e}\\\\ \begin{aligned}2\cos 50^{\circ}&\cos 40^{\circ}-2\sin 95^{\circ}\sin 85^{\circ}\\ &=\cos \left ( 50^{\circ}+40^{\circ} \right )+\cos \left ( 50^{\circ}-40^{\circ} \right )+\cos \left ( 95^{\circ}+85^{\circ} \right )-\cos \left ( 95^{\circ}-85^{\circ} \right )\\ &=2\cos 90^{\circ}+\cos 10^{\circ}+\cos 180^{\circ}-\cos 10^{\circ}\\ &=-1 \end{aligned}.

\begin{array}{ll}\\ \fbox{14}.&\textrm{Nilai}\: \: \displaystyle \frac{\cos 75^{\circ}-\cos 15^{\circ}}{\sin 75^{\circ}-\sin 15^{\circ}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -1&&\textrm{d}.\quad \displaystyle \frac{1}{3}\sqrt{6}\\\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\sqrt{2}&\textrm{c}.\quad 1&\textrm{e}.\quad \displaystyle \frac{1}{2}\sqrt{6} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{a}\\\\ \begin{aligned}\displaystyle \frac{\cos 75^{\circ}-\cos 15^{\circ}}{\sin 75^{\circ}-\sin 15^{\circ}}&=\displaystyle \frac{-2\sin \displaystyle \frac{1}{2}\left ( 75^{\circ}+15^{\circ} \right )\sin \frac{1}{2}\left ( 75^{\circ}-15^{\circ} \right )}{2\cos \displaystyle \frac{1}{2}\left ( 75^{\circ}+15^{\circ} \right )\sin \frac{1}{2}\left ( 75^{\circ}-15^{\circ} \right )}\\ &=-\displaystyle \frac{2\sin 45^{\circ}\times \sin 30^{\circ}}{2\cos 45^{\circ}\times \sin 30^{\circ}}\\ &=-\tan 45^{\circ}\\ &=-1 \end{aligned}.

\begin{array}{ll}\\ \fbox{15}.&\textrm{Bentuk sederhana dari}\: \: 2\cos \left ( x+\displaystyle \frac{\pi }{4} \right ). \cos \left ( \displaystyle \frac{3}{4}\pi -x \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \cos (2x+\pi )&&\textrm{d}.\quad -\cos \left ( 2x-\displaystyle \frac{\pi }{2} \right )\\\\ \textrm{b}.\quad \cos \left ( 2x+\displaystyle \frac{\pi }{2} \right )&\textrm{c}.\quad \cos 2x&\textrm{e}.\quad \sin 2x-1 \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{e}\\\\ \begin{aligned}2\cos \left ( x+\displaystyle \frac{\pi }{4} \right ). \cos \left ( \displaystyle \frac{3}{4}\pi -x \right )&=\cos \left ( x+\displaystyle \frac{\pi }{4}+\displaystyle \frac{3}{4}\pi -x \right )+\cos \left (x+\displaystyle \frac{\pi }{4} -\left ( \displaystyle \frac{3}{4}\pi -x \right ) \right )\\ &=\cos \pi +\cos \left ( 2x-\frac{2}{4}\pi \right )\\ &=-1+\cos \left ( \displaystyle \frac{1}{2}\pi -2x \right ),\quad\quad \textrm{ingat bahwa}\: \cos (-\alpha )=\cos \alpha \\ &=-1+\sin 2x\\ &=\sin 2x-1 \end{aligned}.

\begin{array}{ll}\\ \fbox{16}.&\textrm{Nilai dari}\: \: \cos 80^{\circ}+\cos 40^{\circ}-\cos 20^{\circ}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 1&&\textrm{d}.\quad -\displaystyle \frac{1}{2}\\\\ \textrm{b}.\quad -1&\textrm{c}.\quad \displaystyle \frac{1}{2}&\textrm{e}.\quad 0\end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{e}\\\\ \begin{aligned}\cos 80^{\circ}+\cos 40^{\circ}-\cos 20^{\circ}&=2\cos \displaystyle \frac{1}{2}\left ( 80^{\circ}+40^{\circ} \right )\cos \displaystyle \frac{1}{2}\left ( 80^{\circ}-40^{\circ} \right )-\cos 20^{\circ}\\ &=2\cos 60^{\circ}\cos 20^{\circ}-\cos 20^{\circ}\\ &=2.\displaystyle \frac{1}{2}.\cos 20^{\circ}-\cos 20^{\circ}\\ &=\cos 20^{\circ}-\cos 20^{\circ}\\ &=0 \end{aligned}.

\begin{array}{ll}\\ \fbox{17}.&\textrm{Bentuk sederhana dari}\: \: 2\sin \left ( \displaystyle \frac{3}{4}\pi +x \right ).\cos \left ( \displaystyle \frac{\pi }{4}+x \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 1-\sin 2x&&\textrm{d}.\quad \cos 2x\\\\ \textrm{b}.\quad 1+\sin 2x&\textrm{c}.\quad 1-\cos 2x&\textrm{e}.\quad -\cos 2x\end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{a}\\\\ \begin{aligned}2\sin \left ( \displaystyle \frac{3}{4}\pi +x \right ).\cos \left ( \displaystyle \frac{\pi }{4}+x \right )&=2\sin \left ( 135^{\circ}+x \right ).\cos \left ( 45^{\circ}+x \right )\\ &=2\sin \left ( 90^{\circ}+45^{\circ}+x \right ).\cos \left ( 45^{\circ}+x \right )\\ &=2\cos \left ( 45^{\circ}+x \right ).\cos \left ( 45^{\circ}+x \right )\\ &=2\cos ^{2}\left ( 45^{\circ}+x \right )\\ &=2\left ( \cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x \right )^{2}\\ &=2\left ( \frac{1}{2}\sqrt{2}.\cos x-\displaystyle \frac{1}{2}\sqrt{2}.\sin x \right )^{2}\\ &=\displaystyle \left ( \cos x-\sin x \right )^{2}\\ &=\cos ^{2}x+\sin ^{2}x-2\sin x\cos x\\ &=1-\sin 2x \end{aligned}.

Sumber Referensi

  1. Kartini, Suprapto, Subandi, dan Untung Setiyadi. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: Intan Pariwara.
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InsyaAllah

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Lanjutan Contoh Soal Persamaan Lingkaran dan Garis Singgung 2

\begin{array}{ll}\\ \fbox{11}.&\textrm{Persamaan garis singgung lingkaran}\: \: x^{2}+y^{2}+8x-3y-24=0,\\ &\textrm{di titik}\: \: (2,4)\: \: \textrm{adalah}....\\ &\textrm{a}.\quad 12x-5y-44=0\\ &\textrm{b}.\quad 12x+5y-44=0\\ &\textrm{c}.\quad 12x-y-50=0\\ &\textrm{d}.\quad 12x+y-50=0\\ &\textrm{e}.\quad 12x+y+50=0 \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{b}\\\\ \begin{aligned}x^{2}+y^{2}+8x-3y-24&=x^{2}+8x+16+y^{2}-3y+\displaystyle \frac{9}{4}-24=16+\frac{9}{4}\\ (x+4)^{2}+(y-\frac{3}{2})^{2}&=16+\frac{9}{4}+24=42\frac{1}{4}\\ \textrm{Persamaan garis singgung lingkar}&\textrm{an lingkaran di titik}\: \: (x_{1},y_{1})\: \: \textrm{adalah}:\\ (x_{1}+4)(x+4)+(y_{1}-\frac{3}{2})(y-\frac{3}{2})&=42\frac{1}{4},\qquad \textrm{untuk}\: \: (x_{1},y_{1})=(2,4),\: \textrm{maka}\\ (2+4)(x+4)+(4-\frac{3}{2})(y-\frac{3}{2})&=\frac{169}{4}\\ 6(x+4)+\frac{5}{2}(y-\frac{3}{2})&=\frac{169}{4}\\ 24(x+4)+5(2y-3)&=169\\ 24x+96+10y-15&=169\\ 24x+10y&=169-96+15=88\\ 12x+5y-44&=0 \end{aligned}.

\begin{array}{ll}\\ \fbox{12}.&\textrm{Sebuah garis singgung}\: \: g\: \: \textrm{menyinggung lingkaran yang berpusat di}\: \: (-2,5)\: \: \textrm{dan}\\ &\textrm{berjari-jari}\: \: 2\sqrt{10}\: \: \textrm{di titk}\: \: (4,3).\: \textrm{Maka persamaan garis singgung}\: \: g\: \: \textrm{adalah}....\\ &\textrm{a}.\quad y=3x+9\\ &\textrm{b}.\quad y=3x-9\\ &\textrm{c}.\quad y=-3x+9\\ &\textrm{d}.\quad y=-3x-9\\ &\textrm{e}.\quad y=3x+21\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{b}\\\\ \begin{aligned}(x-a)^{2}+(y-b)^{2}&=r^{2}\\ &\begin{cases} \textrm{Pusat} & =(-2,5) \\ \textrm{r} & =2\sqrt{10} \end{cases} (x+2)^{2}+(y-5)^{2}&=(2\sqrt{10})^{2}\\ (x_{1}+2)(x+2)+(y_{1}-5)(y-5)&=40,\qquad \textrm{menyingung garis}\: \: g\: \: \textrm{di}\: (4,3)\\ (4+2)(x+2)+(3-5)(y-5)&=40\\ 6x+12-2y+10&=40\\ 6x-2y&=40-12-10\\ 3x-y&=9\\ -y&=-3x+9\\ y&=3x-9 \end{aligned}.

\begin{array}{ll}\\ \fbox{13}.&\textrm{Suatu lingkaran dengan titik pusatnya terletak pada kurva}\: \: y=\sqrt{x}\: \: \textrm{dan melalui}\\ &\textrm{titik asal}\: \: O(0,0).\: \textrm{Jika diketahui absis titik pusat lingkaran tersebut adalah}\: \: a,\\ &\textrm{maka persamaan garis singgung lingkaran yang melalui titik}\: \: O\: \: \textrm{tersebut adalah}....\\ &\textrm{a}.\quad y=-x\\ &\textrm{b}.\quad y=-x\sqrt{a}\\ &\textrm{c}.\quad y=-ax\\ &\textrm{d}.\quad y=-2x\sqrt{2}\\ &\textrm{e}.\quad y=-2ax\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{b}\\\\ \begin{array}{|l|c|l|}\hline \textup{Pusat lingkaran}&\begin{aligned}&\textrm{Gradien garis singgung}\\ &\textrm{yang tegak lurus dengan garis}\\ &\textrm{yang melalui titik pusat}\\ &\textrm{lingkaran yang bergradien}\: \: m_{L} \end{aligned}&\begin{aligned}&\textrm{Persamaan garis singgung}\\ &\textrm{yang melalui titik}\\ &\textrm{asal}\: \: O(0,0) \end{aligned}\\\hline (a,b)=\left ( a,\sqrt{a} \right )&\begin{aligned}m.m_{1}&=-1\\ m.\frac{y}{x}&=-1\\ m&=-\frac{x}{y}\\ m&=-\displaystyle \frac{a}{\sqrt{a}}=-\sqrt{a} \end{aligned}&\begin{aligned}y&=mx,\quad \textrm{karena melalui}\\ y&=-\sqrt{a}x,\: \textrm{titik asal}\\ y&=-x\sqrt{a} \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{14}.&\textrm{Salah satu garis singgung yang bersudut}\: \: 120^{\circ}\: \: \textrm{terhadap sumbu x positif terhadap}\\ &\textrm{lingkaran dengan ujung diameter titik}\: \: (7,6)\: \textrm{dan}\: \: (1,-2)\: \: \textrm{adalah}....\\ &\textrm{a}.\quad y=-x\sqrt{3}+4\sqrt{3}+12\\ &\textrm{b}.\quad y=-x\sqrt{3}-4\sqrt{3}+8\\ &\textrm{c}.\quad y=-x\sqrt{3}+4\sqrt{3}-4\\ &\textrm{d}.\quad y=-x\sqrt{3}-4\sqrt{3}-8\\ &\textrm{e}.\quad y=-x\sqrt{3}+4\sqrt{3}+22\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{a}\\\\ \begin{array}{|c|c|c|}\hline \textrm{Pusat Lingkaran}&\textrm{Gradien Garis Singgung}&\textrm{Jari-jari}\\\hline \begin{aligned}(a,b)&=\left ( \displaystyle \frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2} \right )\\ &=\left ( \displaystyle \frac{7+1}{2},\frac{6+(-2)}{2} \right )\\ &=(4,2)\\ &\\ &\\ \end{aligned}&\begin{aligned}m&=\tan 120^{\circ}\\ &=-\tan \left ( 180^{\circ}-60^{\circ} \right )\\ &=-\tan 60^{\circ}\\ &=-\sqrt{3}\\ &\\ & \end{aligned}&\begin{aligned}r&=\textrm{jarak titik}\\ &\: \: \: \: \: \, \textrm{singgung ke pusat}\\ &=\sqrt{(7-4)^{2}+(6-2)^{2}}\\ &=\sqrt{3^{2}+4^{2}}\\ &=\sqrt{25}\\ &=5 \end{aligned}\\\hline \multicolumn{3}{|c|}{\begin{aligned}\textrm{Sehingga persamaan }&\: \textrm{garis singgungnya adalah:}\\\\ (y-b)&=m(x-a)\pm r\sqrt{1+m^{2}}\\ (y-2)&=-\sqrt{3}(x-4)\pm 5\sqrt{1+(-\sqrt{3})^{2}}\\ y-2&=-\sqrt{3}x+4\sqrt{3}\pm 5\sqrt{1+4}\\ y&=-\sqrt{3}x+4\sqrt{3}+2\pm 10=\begin{cases} -\sqrt{3}x+4\sqrt{3}+2+ 10 \\ -\sqrt{3}x+4\sqrt{3}+2- 10 \end{cases}\\ y&=\begin{cases} -\sqrt{3}x+4\sqrt{3}+12 & \\ -\sqrt{3}x+4\sqrt{3}-8 & \end{cases} \end{aligned}}\\\hline \end{array}.

Perhatikanlah gambar berikut sebagai ilustrasinya

372

\begin{array}{ll}\\ \fbox{15}.&\textrm{Salah satu garis singgung lingkaran}\: \: x^{2}+y^{2}=10\: \: \textrm{yang ditarik dari}\\ &\textrm{titik}\: \: (4,2)\: \: \textrm{adalah}....\\ &\textrm{a}.\quad x+3y=10\\ &\textrm{b}.\quad x-3y=10\\ &\textrm{c}.\quad -x-3y=10\\ &\textrm{d}.\quad 2x+y=10\\ &\textrm{e}.\quad x+2y=10\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{a}\\\\.

\begin{array}{|c|c|}\hline \begin{aligned}&\textrm{Garis Singgung}\\ &\quad\quad \textrm{di titik}\\ &(x_{1},y_{1})=(4,2) \end{aligned}&\begin{aligned}&\textrm{Tahapan menentukan}\\ &\quad\qquad \textrm{harga}\: \: m \end{aligned}\\\hline \begin{aligned}y-y_{1}&=m(x-x_{1})\\ y-2&=m(x-4)\\ y&=mx-4m+2\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}x^{2}+y^{2}&=10\\ x^{2}+\left ( mx-4m+2 \right )^{2}&=10\\ x^{2}+m^{2}x^{2}+16m^{2}+4-8m^{2}x+4mx-16m&=10\\ x^{2}+m^{2}x^{2}+16m^{2}-8m^{2}x+4mx-16m-6&=0\\ (1+m^{2})x^{2}+(4m-8m^{2})x+16m^{2}-16m-6&=0\begin{cases} a & =1+m^{2} \\ b & =4m-8m^{2} \\ c & =16m^{2}-16m-6 \end{cases} \end{aligned}\\\hline \multicolumn{2}{|c|}{\begin{aligned}\textrm{Syarat menyinggung}\: \: D&=0\\ b^{2}-4ac&=0\\ \left ( 4m-8m^{2} \right )^{2}-4\left ( 1+m^{2} \right )\left ( 16m^{2}-16m-6 \right )&=0\\ 16m^{2}-64m^{3}+64m^{4}-64m^{2}+64m+24-64m^{4}+64m^{3}+24m^{2}&=0\\ -24m^{2}+64m+24&=0\\ -3m^{2}+8m+3&=0\\ (m-3)(3m+1)&=0\\ m=3\: \: \textrm{atau}\: \: m&=-\displaystyle \frac{1}{3}\\ m&=\begin{cases} 3 & \Rightarrow y=3x-10\\ &\Rightarrow 3x-y=10\\ -\displaystyle \frac{1}{3} & \Rightarrow y=-\displaystyle \frac{1}{3}x+\frac{4}{3}+2\\ &\Rightarrow x+3y=10 \end{cases} \end{aligned}}\\\hline \end{array}.

Sumber Referensi

  1. Kartini, Suprapto, Subandi, dan Untung Setiyadi. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: Intan Pariwara.
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Contoh Soal Persamaan Lingkaran dan Garis Singgung (Kelas XI)

\begin{array}{ll}\\ \fbox{1}.&\textrm{Persamaan lingkaran yang berpusat di}\: \: P(-2,5)\: \: \textrm{dan melalui titik}\: \: T(3,4)\: \: \textrm{adalah}....\\ &\textrm{a}.\quad (x+2)^{2}+(y-5)^{2}=26\\ &\textrm{b}.\quad (x-3)^{2}+(y+5)^{2}=36\\ &\textrm{c}.\quad (x+2)^{2}+(y-5)^{2}=82\\ &\textrm{d}.\quad (x-3)^{2}+(y+5)^{2}=82\\ &\textrm{e}.\quad (x+2)^{2}+(y+5)^{2}=82\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{a}\\\\ \begin{array}{|l|l|l|}\hline \multicolumn{2}{|c|}{\textrm{Persamaan Lingkaran Berpusat di}\: \: (a,b)\: \: \textrm{adalah}}&\\ \multicolumn{2}{|c|}{(x-a)^{2}+(y-b)^{2}=r^{2}}&\\\cline{1-2} \textrm{Pusat di}\: \: P(-2,5)&\textrm{Melalui Titik}\: \: T(3,4)&\textrm{Sehinga persamaan lingkarannya}\\\cline{1-2} \begin{aligned}(x-a)^{2}+(y-b)^{2}&=r^{2}\\ (x+2)^{2}+(y-5)^{2}&=r^{2}\\ &\\ & \end{aligned}&\begin{aligned}(x-a)^{2}+(y-b)^{2}&=r^{2}\\ (3+2)^{2}+(4-5)^{2}&=r^{2}\\ 5^{2}+(-1)^{2}&=r^{2}\\ 26&=r^{2} \end{aligned}&\begin{aligned}&\textrm{adalah}:\\ &(x+2)^{2}+(y-5)^{2}=r^{2}=26\\ &(x+2)^{2}+(y-5)^{2}=26\\ & \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Koordinat titik pusat dan jari-jari lingkaran}\: \: x^{2}+y^{2}-4x+6y+4=0\: \: \textrm{adalah}....\\ &\textrm{a}.\quad (-3,2)\: \: \textrm{dan}\: \: 3\\ &\textrm{b}.\quad (3,-2)\: \: \textrm{dan}\: \: 3\\ &\textrm{c}.\quad (-2,-3)\: \:\textrm{ dan}\: \: 3\\ &\textrm{d}.\quad (2,-3)\: \: \textrm{dan}\: \: 3\\ &\textrm{e}.\quad (2,3)\: \: \textrm{dan}\: \: 3\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{d}\\\\ \textbf{Alterntif 1}\\\\ \begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\textrm{Persamaan Lingkaran Berpusat di}\: \: (a,b)\: \: \textrm{dan berjari-jari}\: \: r\: \: \textrm{adalah}}\\ \multicolumn{2}{|c|}{\begin{aligned}(x-a)^{2}+(y-b)^{2}&=r^{2}\\ x^{2}+y^{2}-4x+6y+4&=0\\ x^{2}-4x+y^{2}+6y+4&=0\\ x^{2}-4x+4-4+y^{2}+6y+9-9+4&=0\\ (x-2)^{2}-4+(y+3)^{2}-9+4&=0\\ (x-2)^{2}+(y+3)^{2}&=4+9-4\\ (x-2)^{2}+(y+3)^{2}&=9\\ (x-2)^{2}+(y-(-3))^{2}&=3^{2}\begin{cases} \textrm{Pusat} & =(2,-3) \\ \textrm{dan}\\ \: r & = 3 \end{cases} \end{aligned}}\\\cline{1-2} \end{array}\\\\ \textbf{Alternatif 2}\\ \begin{aligned}\textrm{Diketahui}&\: \textrm{persamaan lingkaran}:\: \: x^{2}+y^{2}-4x+6y+4=0\begin{cases} A & =-4 \\ B & =6 \\ C & =4 \end{cases}\\ &x^{2}+y^{2}+Ax+By+C=0\\ &\begin{cases} \textrm{Pusat} & =\left ( -\displaystyle \frac{1}{2}A,\: -\frac{1}{2}B \right )=\left ( -\frac{1}{2}\cdots ,\: -\frac{1}{2}\cdots \right )=(\cdots ,\cdots ) \\ \textrm{Jari-jari} & =\sqrt{\displaystyle \frac{1}{4}A^{2}+\frac{1}{4}B^{2}-C}=\sqrt{\displaystyle \frac{1}{4}\cdots ^{2}+\frac{1}{4}\cdots ^{2}-\cdots }=\sqrt{\cdots } \end{cases} \end{aligned}\\\\ \textrm{Silahkan dilengkapi sendiri}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Suatu lingkaran}\: \: x^{2}+y^{2}-4x+2y+p=0\: \: \textrm{berjari-jari 3, maka nilai}\: \: p\: \: \textrm{adalah}....\\ &\textrm{a}.\quad -1\\ &\textrm{b}.\quad -2\\ &\textrm{c}.\quad -3\\ &\textrm{d}.\quad -4\\ &\textrm{e}.\quad -5\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{d}\\\\ \begin{aligned}r=\sqrt{\displaystyle \frac{A^{2}}{4}+\frac{B^{2}}{4}-C}&=3\\ \displaystyle \sqrt{\frac{(-4)^{2}}{4}+\frac{2^{2}}{4}-p}&=3\\ \displaystyle \frac{16}{4}+\frac{4}{4}-p&=9\\ 4+1-p&=9\\ -p&=9-5\\ p&=-4 \end{aligned}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Diketahui lingkaran}\: \: x^{2}+y^{2}+4x+ky-12=0\: \: \textrm{melalui titik}\: \: (-2,8)\: \: \textrm{maka jari-jari}\\ &\textrm{lingkaran tersebut adalah}....\\ &\textrm{a}.\quad 1\\ &\textrm{b}.\quad 5\\ &\textrm{c}.\quad 6\\ &\textrm{d}.\quad 12\\ &\textrm{e}.\quad 25\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{b}\\\\ \begin{aligned}\textrm{Lingkran} \qquad \qquad\quad\quad &\\ x^{2}+y^{2}+4x+ky-12&=0\: \: \textrm{melalui}\: \: (-2,8)\: \: \textrm{berarti }\\ x^{2}+y^{2}+4x+ky-12&=(-2)^{2}+8^{2}+4(-2)+k.8-12\\ 0&=4+64-8-12+8k\\ 0&=48+8k\\ -6&=k\\ \textrm{Sehingga}\qquad\quad\quad\quad r\: \, &=\sqrt{\displaystyle \frac{4^{2}}{4}+\frac{(-6)^{2}}{4}-(-12)}=\sqrt{\displaystyle 4+9+12}=\sqrt{25}=5\\ \end{aligned}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Persmaan lingkaran}\: \: x^{2}+y^{2}+px+8y+9=0\: \: \textrm{menyinggung sumbu X. Pusat lingkaran tersebut adalah}....\\ &\textrm{a}.\quad (6,-4)\\ &\textrm{b}.\quad (6,6)\\ &\textrm{c}.\quad (3,-4)\\ &\textrm{d}.\quad (-6,-4)\\ &\textrm{e}.\quad (3,4)\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{c}\\\\ \begin{aligned}\textrm{Lingkaran}\: \: x^{2}+y^{2}+px+8y+9&=0\: \: \textrm{menyinggung sumbu X} \: (\textrm{sejajar}\: \: y=0)\\ \textrm{maka,}\quad\quad\qquad\qquad\qquad\quad\quad\quad\: \: \: &\\ y=0\: \Rightarrow \: x^{2}+0^{2}+px+8.0+9&=0\\ x^{2}+px+9&=0\begin{cases} a & =1 \\ b & =p \\ c & =9 \end{cases}\: ,\: \textrm{syarat menyinggung}\: \: D=b^{2}-ac=0\\ b^{2}-4ac&=0\\ p^{2}-4(1)(9)&=0\\ p^{2}-36&=(p+6)(p-6)=0\\ x&=-6\: \: \textrm{atau}\: \: x=6\\ p=-6\: \Rightarrow \: x^{2}+y^{2}-6x+8y+9&=0\Rightarrow \textrm{pusatnya adalah}\: \: \left ( -\displaystyle \frac{A}{2},-\frac{B}{2} \right )=(3,-4)\\ p=6\: \Rightarrow \: x^{2}+y^{2}+6x+8y+9&=0\Rightarrow \textrm{pusatnya adalah}\: \: \left ( -\displaystyle \frac{A}{2},-\frac{B}{2} \right )=(-3,-4) \end{aligned}.

Sebagai ilustrasi perhatikanlah gambar berikut ini

365

\begin{array}{ll}\\ \fbox{6}.&\textrm{Titik-titik berikut yang posisinya berada di luar lingkaran}\: \: x^{2}+y^{2}-2x+8y-32=0\: \: \textrm{adalah}.... \\ &\textrm{a}.\quad (0,0)\\ &\textrm{b}.\quad (-6,-4)\\ &\textrm{c}.\quad (-3,2)\\ &\textrm{d}.\quad (3,1)\\ &\textrm{e}.\quad (4,1)\\\end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|c|c|l|c|}\hline \textrm{Opsi}&\textrm{Titik}&\textrm{Lingkaran}\equiv 0^{2}+0^{2}-2.0+8.0-32=-32&\textrm{Keterangan}\\\hline \textrm{a}&(0,0)&0^{2}+0^{2}-2.0+8.0-32=-32&\textrm{dalam}\\\hline \textrm{b}&(-6,-4)&(-6)^{2}+(-4)^{2}-2(-6)+8(-4)-32=0&\textrm{pada}\\\hline \textcircled{c}&(-3,2)&(-3)^{2}+(2)^{2}-2(-3)+8(2)-32=3&\textbf{di luar}\\\hline \textrm{d}&(3,1)&3^{2}+1^{2}-2.3+8.1-32=-20&\textrm{dalam}\\\hline \textrm{e}&(4,1)&4^{2}+1^{2}-2.4+8.1-32=-15&\textrm{dalam}\\\hline \end{array}.

Sebagai ilustrasi perhatikanlah gambar berikut:

366

\begin{array}{ll}\\ \fbox{7}.&\textrm{Diketahui garis}\: \: x-2y=5\: \: \textrm{memotong lingkaran}\: \: x^{2}+y^{2}-4y+8y+10=0\: \: \textrm{di titik A dan B}.\\ &\textrm{Panjang ruas garis AB adalah}....\\ &\textrm{a}.\quad 4\sqrt{2}\\ &\textrm{b}.\quad 2\sqrt{5}\\ &\textrm{c}.\quad \sqrt{10}\\ &\textrm{d}.\quad 5\\ &\textrm{e}.\quad 4\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{b}\\\\ \begin{aligned}\textrm{Perhatikanlah bahwa},\quad\quad\qquad\qquad\quad\quad\, &\\ x^{2}+y^{2}-4x+8y+10&=0,\quad \textrm{dengan}\: \: x=2y+5\\ (2y+5)^{2}+y^{2}-4(2y+5)+8y+10&=0\\ 4y^{2}+20y+25+y^{2}-8y-20+8y+10&=0\\ 5y^{2}+20y+15&=0\\ y^{2}+4y+3&=0\\ (y+1)(y+3)&=0\\ y=-1\: \: \vee \: \: y&=-3\\ \textrm{untuk nilai};\qquad y&=-3\Rightarrow x=2(-3)+5=-1,\quad A(-1,-3)\\ \qquad y&=-1\Rightarrow x=2(-1)+5=3,\qquad B(3,-1)\\ \textrm{maka},\qquad \textrm{AB}&=\sqrt{(3-(-1))^{2}+(-1-(-3))^{2}}\\ &=\sqrt{4^{2}+2^{2}}\\ &=\sqrt{16+4}\\ &=\sqrt{20}\\ &=2\sqrt{5} \end{aligned}.

Berikut ilustrasi gambarnya;

367

\begin{array}{ll}\\ \fbox{8}.&\textrm{Kekhususan persamaan lingkaran}\: \: x^{2}+y^{2}-6x-6y+6=0\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \textrm{menyinggung sumbu X}\\ &\textrm{b}.\quad \textrm{menyinggung sumbu Y}\\ &\textrm{c}.\quad \textrm{berpusat di}\: \: O(0,0)\\ &\textrm{d}.\quad \textrm{titik pusatnya terletak pada}\: \: x-y=0\\ &\textrm{e}.\quad \textrm{berjari-jari 3}\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{d}\\\\ \begin{array}{|c|l|c|}\hline \multicolumn{3}{|c|}{\begin{aligned}x^{2}+y^{2}-6x-6y+6&=0\\ x^{2}-6x+9+y^{2}-6y+9+6&=9+9\\ (x-3)^{2}+(y-3)^{2}&=18-6\\ (x-3)^{2}+(y-3)^{2}&=12\\ (x-3)^{2}+(y-3)^{2}&=\left ( 2\sqrt{3} \right )^{2}\\ \textrm{lingkaran ini}&\begin{cases} \textrm{Pusat} &=(3,3) \\ \textrm{Jari-jari} & =2\sqrt{3} \end{cases} \end{aligned}}\\\hline \textrm{Opsi}&\textrm{Pernyataan}&\textrm{Keterangan}\\\hline \textrm{a}&\textrm{menyinggung sumbu X}&\textrm{tidak tepat}\\\hline \textrm{b}&\textrm{menyinggung sumbu Y}&\textrm{tidak tepat}\\\hline \textrm{c}&\textrm{berpusat di}\: \: O(0,0)&\textrm{tidak tepat}\\\hline \textcircled{d}&\textrm{titik pusatnya terletak pada garis}\: \: x-y=0&\textbf{tepat}\\\hline \textrm{e}&\textrm{berjari-jari 3}&\textrm{tidak tepat}\\\hline \end{array}.

Berikut ilustrasinya,

368

\begin{array}{ll}\\ \fbox{9}.&\textrm{Lingkaran}\: \: x^{2}+y^{2}+2ax+2by+c=0\: \: \textrm{menyinggung sumbu Y jika}\: \: c\: =....\\ &\textrm{a}.\quad ab\\ &\textrm{b}.\quad ab^{2}\\ &\textrm{c}.\quad a^{2}b\\ &\textrm{d}.\quad a^{2}\\ &\textrm{e}.\quad b^{2} \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{e}\\\\ \begin{aligned}x^{2}+y^{2}+2ax+2by+c&=0\\ x=0\Rightarrow 0^{2}+y^{2}+2a.0+2by+c&=0\\ y^{2}+2by+c&=0\begin{cases} a & =1 \\ b & =2b \\ c & =c \end{cases}\\ \textrm{Syarat menyinggung}&\: \textrm{adalah}:\\ D=b^{2}-4ac&=0\\ (2b)^{2}-4.1.c&=0\\ 4c&=4b^{2}\\ c&=b^{2} \end{aligned}.

\begin{array}{ll}\\ \fbox{10}.&\textrm{Diketahui pusat lingkaran L terletak dikuadran I dan berada di sepanjang garis}\: \: y=2x.\\ &\textrm{Jika lingkaran L menyinggung sumbu Y di titik}\: \: (0,6),\: \textrm{maka persamaan lingkaran L adalah}....\\ &\textrm{a}.\quad x^{2}+y^{2}-3x-6y=0\\ &\textrm{b}.\quad x^{2}+y^{2}+6x+12y-108=0\\ &\textrm{c}.\quad x^{2}+y^{2}+12x+6y-72=0\\ &\textrm{d}.\quad x^{2}+y^{2}-12x-6y=0\\ &\textrm{e}.\quad x^{2}+y^{2}-6x-12y+36=0 \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{e}\\\\ \begin{aligned}(x-a)^{2}+(y-b)^{2}=r^{2},\quad \: \, \, &\\ \textrm{menyinggung titik}\: \: (0,6)\Rightarrow \, &\textrm{berarti pusat lingkaran L juga terletak pada garis}\: \: y=6.\\ \textrm{ini berarti pusat lingkaran}\: \: \, &\: \textrm{L berpusat di}\: \: (x,2x)=(\frac{y}{2},y),\: \: \textrm{dengan}\: \: y=6.\: \textrm{Sehingga pusatnya berada pada titik}\: \: (3,6).\\ \textrm{Maka persamaan lingkaran}\: \, &\textrm{adalah}\: \: (x-3)^{2}+(y-6)^{2}=3^{2}\: \: \textrm{ingat}\: \: r=\textrm{absis}\: \: x=3\\ (x-3)^{2}+(y-6)^{2}&=x^{2}-6x+9+y^{2}+12x+36=9\\ &\Leftrightarrow \, x^{2}+y^{2}-6x+12y+36=0 \end{aligned}.

Berikut ilustrasi gambarnya

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Sumber Referensi

  1. Kartini, Suprapto, Subandi, dan Untung Setiyadi. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: Intan Pariwara.
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Lingkaran (KTSP MA/SMA Kelas XI)

A. Persamaan Lingkaran dan Garis Singgung

\begin{array}{|l|c|c|c|}\hline &\multicolumn{3}{c|}{\textbf{Persamaan}}\\\cline{2-4} \raisebox{1.5ex}[0cm][0cm]{\textrm{Lingkaran}} &x^{2}+y^{2}=r^{2}&(x-p)^{2}+(y-q)^{2}=r^{2}&x^{2}+y^{2}+Ax+By+C=0\\\hline \textrm{Pusat}&(0,0)&(p,q)&\left ( -\frac{1}{2}A,-\frac{1}{2}B \right )\\\hline \textrm{Jari-jari}&r&r&r=\sqrt{\displaystyle \frac{1}{4}\left ( A^{2}+B^{2} \right )-C}\\\hline \begin{aligned}&\textrm{Pesamaan garis}\\ &\textrm{singgung melalui}\\ &\textrm{titik}\: \: (x_{1},y_{1})\\ &\textrm{pada lingkaran} \end{aligned}&x_{1}x+y_{1}y=r^{2}&\begin{aligned}&(x_{1}-p)(x-p)\\ &\: +(y_{1}-q)(y-q)=r^{2} \end{aligned}&\begin{aligned}&x_{1}x+y_{1}y\\ &\: +\displaystyle \frac{A}{2}(x_{1}+x)\\ &\: +\displaystyle \frac{B}{2}(y_{1}+y)+C=0 \end{aligned}\\\hline \begin{aligned}&\textrm{Persamaan garis}\\ &\textrm{singgung dengan}\\ &\textrm{gradien}\: \: m \end{aligned}&\begin{aligned}&y=mx\\ &\: \pm r\sqrt{m^{2}+1} \end{aligned}&\begin{aligned}&(y-q)=m(x-a)\\ &\: \pm r\sqrt{m^{2}+1} \end{aligned}&\begin{aligned}&y+\frac{1}{2}B=m(x+\frac{1}{2}A)\\ &\: \pm \sqrt{\displaystyle \frac{1}{4}\left ( A^{2}+B^{2} \right )-C}.\sqrt{m^{2}+1} \end{aligned}\\\hline \end{array}.

B. Posisi Titik dan Garis terhadap Lingkaran

\begin{array}{|l|l|l|c|c|}\hline &&&&k=\textrm{Kuasa titik}\: \: M(a,b)\\ \textrm{Titik}&\textrm{Posisi}&x^{2}+y^{2}=r^{2}&(x-p)^{2}+(y-q)^{2}=r^{2}&\textrm{terhadap lingkaran}\\ &&&&x^{2}+y^{2}+Ax+By+C=0\\\hline &\textrm{dalam}&a^{2}+b^{2}<r^{2}&(a-p)^{2}+(a-q)^{2}<r^{2}&k=a^{2}+b^{2}+Aa+Bb+C<0\\\cline{2-5} \textrm{M(a,b)}&\textrm{pada}&a^{2}+b^{2}=r^{2}&(a-p)^{2}+(a-q)^{2}=r^{2}&k=a^{2}+b^{2}+Aa+Bb+C=0\\\cline{2-5} &\textrm{Luar}&a^{2}+b^{2}>r^{2}&(a-p)^{2}+(a-q)^{2}>r^{2}&k=a^{2}+b^{2}+Aa+Bb+C>0\\\hline \textrm{Garis}&\multicolumn{4}{c|}{\begin{cases} \bullet &\textrm{memotong lingkaran di dua titik}\: \: (D>0)\: \textrm{ada garis dan titik polar} \\ \bullet &\textrm{menyinggung lingkaran}\: \: (D=0) \\ \bullet &\textrm{tidak memotong ataupun menyinggung}\: \: (D<0) \end{cases}}\\\hline \multicolumn{4}{|c|}{\textrm{Jarak Titik}\: \: M(a,b)\: \: \textrm{terhadap lingkaran berpusat di}\: \: P(p,q)}&\left | MP \right |=r=\left | \displaystyle \frac{Ap+Bq+C}{\sqrt{A^{2}+B^{2}}} \right |\\\hline \end{array}.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ 1.&\textrm{Sebuah lingkaran yang berpusat pada pangkal koordinat}.\\ &\textrm{a}.\quad \textrm{Tentukanlah persamaan lingkaran yang berjari-jari 5}\\ &\textrm{b}.\quad \textrm{Gambarlah lingkaran (pada soal a.) pada kertas grafiks}\\ &\textrm{c}.\quad \textrm{Lukislah titik-titik dari},\: A(2,3),\: B(4,3),\: \: \textrm{dan}\: \: C(3,6).\\ &\textrm{d}.\quad \textrm{Nyatakan kedudukan titik-titik}\: A,\: B,\: \textrm{dan}\: C\: \textrm{terhadap lingkaran. Di dalam, pada,}\\ &\quad\: \: \: \, \textrm{atau beradakah di luar lingkaran}\end{array}.

Jawab:

Perhatikanlah ilustrasi gambar berikut!

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\begin{array}{|l|l|l|l|}\hline \textrm{a}.&\textrm{Pusat lingkaran}&\textrm{b}.&\textrm{Gambar pada kertas grafik}\\\cline{2-2}\cline{4-4} &\begin{aligned}&x^{2}+y^{2}=5^{2}\\ &\qquad\qquad \updownarrow\\ &x^{2}+y^{2}=25\\ &\textrm{atau}\\ &L\equiv \left \{ (x,y)|x^{2}+y^{2}=25 \right \} \end{aligned}&&\quad\qquad \textrm{Perhatikanlah gambar di atas}\\\hline \textrm{c}.&\textrm{Titik-titik}\: \: A,\: B,\: \textrm{dan}\: \: C&\textrm{d}.&\textrm{Kedudukan titik-titik}\: \: A,\: B,\: \textrm{dan}\: \: C\\\cline{2-2}\cline{4-4} & \textrm{perhatikan gambar di atas}&&\begin{matrix} \bullet \quad \textrm{Titik}\: \: A(2,3)\: \textrm{berada di dalam lingkaran}\\ \bullet \quad \textrm{Titik}\: \: A(4,3)\: \textrm{berada pada lingkaran}\: \: \: \: \: \: \: \\ \bullet \quad \textrm{Titik}\: \: A(3,6)\: \textrm{berada di luar lingkaran}\: \: \: \, \end{matrix}\\\hline \end{array}.

\begin{array}{ll}\\ 2.&\textrm{Tentukanlah persamaan lingkaran yang berpusat di pangkal koordinat}.\\ &\textrm{dan melalui titik}\: \: P(5,-3)\end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}\textrm{Diketahui}&\: \textrm{pusat lingkaran di pangkal koordinat}\: \: O(0,0)\\ \textrm{serta\, ling}&\textrm{karan melalui titik}\: \: P(5,-3),\: \textrm{maka}\\ r&=\sqrt{(x_{p}-0)^{2}+(y_{p}-0)^{2}}\\ &=\sqrt{5^{2}+(-3)^{2}}\\ &=\sqrt{25+9}\\ &=\sqrt{34}\\ \textrm{Sehingga }&,\: \textrm{persamaan lingkarannya adalah}\\ L&\equiv x^{2}+y^{2}=r^{2}\Leftrightarrow x^{2}+y^{2}=34 \end{aligned}.

\begin{array}{ll}\\ 3.&\textrm{Tentukanlah persamaan lingkaran yang berpusat di pangkal koordinat}.\\ &\textrm{dan menyinggung}\: \: k\equiv 2x+y-5=0\end{array}.

Jawab:

Perhatikanlah ilustrasi gambar berikut ini

361          menjadi

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\begin{aligned}\textrm{Diketahui}&\: \textrm{bahwa titik}\: \: O\: \: \textrm{ke garis}\: \: k\: \: \textrm{adalah}\\ r=OA&=\displaystyle \left |\frac{ax_{1}+by_{1}+c}{\sqrt{a^{2}+b^{2}}} \right |\\ &=\displaystyle \left | \frac{2(0)+(0)-5}{\sqrt{2^{2}+1^{2}}} \right |\\ &=\displaystyle \left | \frac{-5}{\sqrt{5}} \right |\\ &=\left | -\sqrt{5} \right |\\ &=-(-\sqrt{5})=\sqrt{5}\qquad \textrm{(ingat, nilai mutlak bilangan negatif adalah bilngan positif)}\end{aligned}\\\\\\ \textrm{Sehingga persamaan lingkarannya adalah}:\\ \qquad L\equiv x^{2}+y^{2}=r^{2}\Leftrightarrow x^{2}+y^{2}=5.

\begin{array}{ll}\\ 4.&\textrm{Tentukanlah pusat dan jari-jari lingkaran berikut?}\\ &\textrm{a}.\quad L\equiv (x+1)^{2}+(y+2)^{2}=9\\ &\textrm{b}.\quad L\equiv (x+1)^{2}+(y-2)^{2}=9\\ &\textrm{c}.\quad L\equiv (x-1)^{2}+(y+2)^{2}=9\\ &\textrm{d}.\quad L\equiv (x-1)^{2}+(y-2)^{2}=9\\ &\textrm{e}.\quad L\equiv (x+3)^{2}+(y-3)^{2}=9\\ &\textrm{f}.\quad L\equiv (x-1)^{2}+(y-2)^{2}=25\\ &\textrm{g}.\quad L\equiv (x-1)^{2}+y^{2}=27\\ &\textrm{h}.\quad L\equiv x^{2}+(y-1)^{2}=27 \end{array}\\\\\\ \textrm{Jawab}:\\\\ L\equiv (x+1)^{2}+(y+2)^{2}=9,\: \: \textrm{pusat di}\: \: \textrm{dan jari-jarinya adalah}\: \: \sqrt{9}=3\\\\ \textrm{Soal yang belum dibahas silahkan diselesaikan sendiri sebagai latihan}.

\begin{array}{ll}\\ 5.&\textrm{Tentukanlah persamaan lingkaran yang berpusat di}\: \: A(2,-1)\: \: \textrm{dan menginggung garis}\\ &4y+3x-12=0\: \: \textrm{di titik}\: \: P\end{array}.

Jawab:

Perhatikanlah ilustrasi gambar berikut!

362

Sehingga

\begin{aligned}&r=AP=\left | \frac{3(2)+4(1)-12}{\sqrt{3^{2}+4^{2}}} \right |=\left | \frac{-10}{5} \right |=\left | -2 \right |=2\\\\ &\textrm{Sehingga persamaan lingkarannya adalah}\: \: L\equiv (x-2)^{2}+(y+1)^{2}=4 \end{aligned}.

\begin{array}{ll}\\ 6.&\textrm{Tentukanlah pusat dan jari-jari dari persamaan lingkaran}\: \: L\equiv 2x^{2}+2y^{2}-2x+6y-3=0\end{array} \\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}&\textrm{Persamaan lingkaran}\: \: L\equiv 2x^{2}+2y^{2}-2x+6y-3=0\Rightarrow x^{2}+y^{2}-x+3y-\displaystyle \frac{3}{2}=0\begin{cases} A & =-1 \\ B & =3 \\ C & =-\displaystyle \frac{3}{2} \end{cases}\\ &\textrm{maka}\: \: \begin{cases} \textrm{Pusat} & =\left ( -\displaystyle \frac{-1}{2},- \frac{3}{2}\right )=\left ( \displaystyle \frac{1}{2},-\frac{3}{2} \right ) \\ \textrm{Jari-jari} & =r=\sqrt{\displaystyle \frac{(-1)^{2}}{4}+\frac{3^{2}}{4}-\left ( -\frac{3}{2} \right )}=\sqrt{\displaystyle \frac{1}{4}+\frac{9}{4}+\frac{6}{4}}=\sqrt{4}=2 \end{cases}\\ &\textrm{Jadi, lingkaran}\: \: 2x^{2}+2y^{2}-2x+6y-3=0\: \: \textrm{berpusat di} \: \: \left ( \displaystyle \frac{1}{2},-\frac{3}{2} \right )\: \: \textrm{dan berjari-jari}\: \: 2\end{aligned}.

\begin{array}{ll}\\ 7.&\textrm{Diketahui lingkaran}\: \: L\equiv 2x^{2}+2y^{2}-4x+3py-30=0\: \: \textrm{dan melalui titik}\: \: (-2,1).\\ &\textrm{Tentukanlah persamaan linkaran baru yang kosentris(sepusat) dan panjang jari-jarinya}\\ &\textrm{dua kali panjang jari-jari lingkaran semula?}\end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|c|c|c|c|}\hline \textrm{Lingkaran semula}&\textrm{Melalui titik}&\textrm{Persamaan lingkaran menjadi}&\textrm{Pusat dan jari-jari}\\\hline 2x^{2}+2y^{2}-4x+3py-30=0&(-2,1)&&\\\cline{1-2} \multicolumn{2}{|l|}{\begin{aligned}\textrm{Proses}&:\\ \: \: &2(-2)^{2}+2(1)^{2}-4(-2)+3p(1)-30=0\\ \Leftrightarrow \: \: &8+2+8+3p-30=0\\ \Leftrightarrow \: \: &3p=12\\ \Leftrightarrow \: \: &p=4 \end{aligned}}&\begin{aligned}&2x^{2}+2y^{2}-4x+12y-30=0\\ &\Leftrightarrow \: \: x^{2}+y^{2}-2x+6y-15=0\\ &\\ & \end{aligned}&\begin{aligned}&\begin{cases} \textrm{Pusat}: \\ \left ( -\displaystyle \frac{1}{2}A,-\frac{1}{2}B \right )\\ =\left ( -\displaystyle \frac{1}{2}.(-2),-\frac{1}{2},6 \right )\\ =(1,-3) \\\\ \textrm{Jari-jari }:\\ \begin{aligned}r&=\sqrt{\left ( -\frac{1}{2}A \right )^{2}+\left ( -\frac{1}{2}B \right )-C}\\ &=\sqrt{1^{2}+(-3)^{2}-(-15)}\\ &=\sqrt{1+9+15}=5 \end{aligned} \end{cases} \\ & \end{aligned}\\\hline \multicolumn{4}{|c|}{\begin{aligned}\textrm{Persamaan}\: &\textrm{lingkaran baru }\\ &\textrm{dengan pusat}\: \: (1,-3)\: \: \textrm{dan jari-jari}\: \: r_{\textrm{baru}}=2r=2.5=10\\ &(x-1)^{2}+(y+3)^{2}=(10)^{2}\\ \Leftrightarrow \: \: &x^{2}-2x+1+y^{2}+6x+9=100\\ \Leftrightarrow \: \: &x^{2}+y^{2}-2x+6y-90=0\end{aligned}}\\\hline \end{array}.

Berikut gambar sebagai ilustrasinya:

363

\begin{array}{ll}\\ 8.&\textrm{Tentukanlah nilai}\: \: p\: \: \textrm{supaya lingkaran}\: \: x^{2}+y^{2}-px-10y+4=0 \\ &\textrm{a}.\quad \textrm{menyinggung sumbu x}\\ &\textrm{b}.\quad \textrm{memotong sumbu x di dua titik}\\ &\textrm{c}.\quad \textrm{tidak memotong dan tidak menyinggung sumbu x} \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|l|l|l|}\hline \multicolumn{3}{|c|}{\begin{aligned}\\ \bullet \: \: \textrm{Lingkaran}:&\\ &x^{2}+y^{2}-px-10y+4=0\\ \textrm{terhadap }\: \: &\textrm{sumbu x},\: \: \textrm{dan}\: \: y=0\\ \textrm{adalah gar}&\textrm{is yang sejajar sumbu x, maka}\\ y=0\Rightarrow \: \: &x^{2}+y^{2}-px-10y+4=0\\ \Rightarrow \: \: &x^{2}-px+4=0\\ & \end{aligned}}\\\hline \textrm{Menyinggung}&\textrm{memotong}&\textrm{Tidak keduanya}\\\hline \begin{aligned}D&= b^{2}-4ac=0\\ &\Leftrightarrow p^{2}-4.1.4=0\\ &\Leftrightarrow p^{2}=16\\ &\Leftrightarrow p=\pm 4\\ & \end{aligned}&\begin{aligned}D&>0\\ &\Leftrightarrow b^{2}-4ac>0\\ &\Leftrightarrow p^{2}-16>0\\ &\Leftrightarrow (p+4)(p-4)>0\\ &\therefore \quad p<-4\: \: \textrm{atau}\: \: p>4 \end{aligned}&\begin{aligned}D&<0\\ &\Leftrightarrow b^{2}-4ac<0\\ &\Leftrightarrow p^{2}-16<0\\ &\Leftrightarrow (p+4)(p-4)<0\\ &\therefore \quad -4<p<4 \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ 9.&\textrm{Tentukanlah nilai}\: \: a\: \: \textrm{supaya lingkaran}\: \: x^{2}+y^{2}=1\: \: \textrm{dan garis}\: \: y=ax+2 \\ &\textrm{a}.\quad \textrm{bersinggungan}\\ &\textrm{b}.\quad \textrm{berpotongan}\\ &\textrm{c}.\quad \textrm{tidak berpotongan maupun bersinggungan} \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}&\textrm{di sini yang kita bahas adalah yang poin b, yaitu untuk}\: \: y=ax+2,\: \: \textrm{maka}\\ &x^{2}+y^{2}=1\\ &x^{2}+(ax+2)^{2}=1\\ &x^{2}+a^{2}x^{2}+4ax+4=1\\ &(1+a^{2})x^{2}+4ax+3=0\\ &\textrm{syarat berpotongan}\: \: D=b^{2}-4ac\geq 0\: \: (\textrm{artinya bersinggungan sekaligus berpotongan di 2 titik})\\ &(4a)^{2}-4(1+a^{2})(3)\geq 0\\ &16a^{2}-12a^{2}-12\geq 0\\ &4a^{2}-12\geq 0\\ &a^{2}-3\geq 0\\ &(a+\sqrt{3})(a-\sqrt{3})\geq 0\\ \therefore \: \: \: \: &a\leq -\sqrt{3}\: \: \textrm{atau}\: \: a\geq \sqrt{3} \end{aligned}.

\begin{array}{ll}\\ 10.&\textrm{Tentukanlah persamaan garis singgung lingkaran}\: \: x^{2}+y^{2}=12\: \: \textrm{dan melalui titik}\: \: P(0,4). \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|c|c|c|}\hline \textrm{Persamaan lingkaran}&\textrm{Persamaan garis singgung di titik}\: \: (x_{1},y_{1})&\textrm{Titik}\: \: (x_{1},y_{1})\: \: \textrm{pada lingkaran}\\\hline x^{2}+y^{2}=12&\begin{aligned}x^{2}+y^{2}&=12\\ xx+yy&=12\\ x_{1}x+y_{1}y&=12\\ \textrm{garis ini melalui}&\: \: \textrm{titik}\\ P(0,4)&, \textrm{maka}\\ x_{1}.0+y_{1}.4&=12\\ y_{1}&=3\: ......\textcircled{1}\end{aligned}&x_{1}^{2}+y_{1}^{2}=12\: ......\textcircled{2}\\\cline{2-3} &\multicolumn{2}{|c|}{\begin{aligned}\textrm{Dari persamaan}&\: \: \textcircled{1}\: \: \textrm{dan}\: \: \textcircled{2}\: \: \textrm{diperoleh}\\ &x_{1}^{2}+y_{1}^{2}=12\\ y_{1}=3\Rightarrow \: \: &x_{1}^{2}+(3)^{2}=12\\ \Leftrightarrow \: \: &x_{1}^{2}+9=12\\ \Leftrightarrow \: \: &x_{1}^{2}=3\\ \Leftrightarrow \: \: &x_{1}=\pm \sqrt{3}\\ \textrm{Sehingga persa}&\textrm{maan garis singgungnya}\left ( x_{1}x+y_{1}y=12 \right )\: \: \textrm{adalah}:\\ &\begin{cases} \textrm{di titik} & (x_{1},y_{1})=(\sqrt{3},3)\: \: \, \, \Rightarrow \sqrt{3}x+3y=12\\ \textrm{di titik} & (x_{1},y_{1})=(-\sqrt{3},3)\Rightarrow -\sqrt{3}x+3y=12 \end{cases}\\ &\end{aligned}}\\\hline \end{array}.

\begin{array}{ll}\\ 10.&\textrm{Tentukanlah persamaan garis singgung lingkaran}\: \: x^{2}+y^{2}=12\: \: \textrm{dan melalui titik}\: \: P(0,4). \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|c|c|c|}\hline \textrm{Persamaan lingkaran}&\textrm{Persamaan garis singgung di titik}\: \: (x_{1},y_{1})&\textrm{Titik}\: \: (x_{1},y_{1})\: \: \textrm{pada lingkaran}\\\hline &\begin{aligned}x^{2}+y^{2}&=12\\ xx+yy&=12\\ x_{1}x+y_{1}y&=12\\ \textrm{garis ini melalui}&\: \: \textrm{titik}\\ P(0,4)&, \textrm{maka}\\ x_{1}.0+y_{1}.4&=12\\ y_{1}&=3\: ......\textcircled{1}\end{aligned}&x_{1}^{2}+y_{1}^{2}=12\: ......\textcircled{2}\\\cline{2-3} x^{2}+y^{2}=12&\multicolumn{2}{|c|}{\begin{aligned}\textrm{Dari persamaan}&\: \: \textcircled{1}\: \: \textrm{dan}\: \: \textcircled{2}\: \: \textrm{diperoleh}\\ &x_{1}^{2}+y_{1}^{2}=12\\ y_{1}=3\Rightarrow \: \: &x_{1}^{2}+(3)^{2}=12\\ \Leftrightarrow \: \: &x_{1}^{2}+9=12\\ \Leftrightarrow \: \: &x_{1}^{2}=3\\ \Leftrightarrow \: \: &x_{1}=\pm \sqrt{3}\\ \textrm{Sehingga persa}&\textrm{maan garis singgungnya}\left ( x_{1}x+y_{1}y=12 \right )\: \: \textrm{adalah}:\\ &\begin{cases} \textrm{di titik} & (x_{1},y_{1})=(\sqrt{3},3)\: \: \, \, \Rightarrow \sqrt{3}x+3y=12\\ \textrm{di titik} & (x_{1},y_{1})=(-\sqrt{3},3)\Rightarrow -\sqrt{3}x+3y=12 \end{cases}\\ &\end{aligned}}\\\hline \end{array}.

Berikut simulasi gambarnya

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Sumber Referensi

  1. Sobirin. 2006. Kompas Matematika Strategi Praktis Menguasai Tes Matematika. Jakarta: Kawan Pustaka.
  2. Wirodikromo, Sartono. 2007. Matematika Jilid 2 IPA untuk Kelas XI. Jakarta: Erlangga.
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Sekilas

Assalamu ‘alaikum Warahmatullahi Wabarakatuhu.

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