Contoh 1 Soal dan Pembahasan Materi Lingkaran

 $\begin{array}{l}\\ 1.& \text{Jari-jari lingkaran dengan persamaan}{x}^2+y^2=48\\ & \text{adalah}....\\ & \text{A}.{\displaystyle 3\sqrt{5}}\\ & \text{B}.4\sqrt{3}\\ & \text{C}.5\sqrt{2}\\ & \text{D}.{\displaystyle 6\sqrt{3}}\\ & \text{E}.7\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}{r}^2& =48\\ r& =\sqrt{48}\\ & =\sqrt{16.3}\\ & =4\sqrt{3}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Titik pusat lingkaran}(x-7{)}^2+(y+9{)}^2=48\\ & \text{adalah}....\\ & \text{A}.{\displaystyle (-7,-9)}\\ & \text{B}.(-7,9)\\ & \text{C}.(7,-9)\\ & \text{D}.{\displaystyle (7,6)}\\ & \text{E}.(15,48)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Jelas bahwa}(a,b)& =(-6,9)\end{array}\end{array}$.

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

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

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