Contoh Soal 1 Materi Lingkaran

 $\begin{array}{ll}\\ 1.&\textrm{Jari-jari lingkaran dengan persamaan}\: \: x^{2}+y^{2}=48\\ &\textrm{adalah}....\\ &\textrm{A}.\quad \displaystyle 3\sqrt{5}\\ &\textrm{B}.\quad \color{red}4\sqrt{3}\\ &\textrm{C}.\quad 5\sqrt{2}\\ &\textrm{D}.\quad \displaystyle 6\sqrt{3}\\ &\textrm{E}.\quad 7\\\\ &\textbf{Jawab}:\qquad \\ &\begin{aligned}r^{2}&=48\\ r&=\sqrt{48}\\ &=\sqrt{16.3}\\ &=4\sqrt{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Titik pusat lingkaran}\: \: (x-7)^{2}+(y+9)^{2}=48\\ &\textrm{adalah}....\\ &\textrm{A}.\quad \displaystyle (-7,-9)\\ &\textrm{B}.\quad (-7,9)\\ &\textrm{C}.\quad \color{red}(7,-9)\\ &\textrm{D}.\quad \displaystyle (7,6)\\ &\textrm{E}.\quad (15,48)\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Jelas bahwa}\: \: \: (a,b)&=(-6,9) \end{aligned} \end{array}$.

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

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

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