Contoh 2 Soal dan Pembahasan Materi Lingkaran

 $\begin{array}{l}\\ 6.& \text{Diketahui lingkaran}{x}^2+y^2+4x+ky-12=0\\ & \text{melalui titik}(-2,8)\text{maka jari-jari lingkaran}\\ & \text{tersebut adalah}....\\ & \text{A}.1\\ & \text{B}.5\\ & \text{C}.6\\ & \text{D}.12\\ & \text{E}.25\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui ingkaran berpusat di}(-2,-{\displaystyle \frac{1}{2}k}),\\ & \text{yaitu}:\\ & x^2+y^2+4x+ky-12=0\\ & \text{melalui}(-2,8)\text{berarti }\\ & (-2{)}^2+8^2+4(-2)+k.8-12=0\\ & 4+64-8-12+8k=0\\ & 48+8k=0\\ & k=-6\\ & \text{Sehingga}r=\sqrt{{\displaystyle \frac{4^2}{4}+\frac{(-6{)}^2}{4}-(-12)}}\\ & =\sqrt{{\displaystyle 4+9+12}}=\sqrt{25}=5\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Persmaan lingkaran}{x}^2+y^2+px+8y+9=0\\ & \text{menyinggung sumbu X. Pusat lingkaran tersebut }\\ & \text{adalah}....\\ & \text{A}.(6,-4)\\ & \text{B}.(6,6)\\ & \text{C}.(3,-4)\\ & \text{D}.(-6,-4)\\ & \text{E}.(3,4)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \mathbf{\text{Lingkaran}}{x}^2+y^2+px+8y+9=0\\ & \text{maka,}\\ & x^2+px+y^2+8y+9=0\\ & {(x+{\displaystyle \frac{1}{2}p})}^2-{\displaystyle \frac{1}{4}{p}^2+(y+4{)}^2-16+9=0}\\ & \iff {(x+{\displaystyle \frac{1}{2}p})}^2+(y+4{)}^2=7+{\displaystyle \frac{1}{4}{p}^2}\\ & \text{karena menyinggung sumbu-X,}R=b=4,\\ & \text{sehingga}\\ & 7+{\displaystyle \frac{1}{4}{p}^2=4^2}\\ & \iff {\displaystyle \frac{1}{4}{p}^2=16-7=9\iff p^2=36\iff p=\pm 6}\\ & p=-6\Rightarrow x^2+y^2-6x+8y+9=0\\ & \Rightarrow \text{pusatnya adalah}(-{\displaystyle \frac{A}{2},-\frac{B}{2}})=(3,-4)\\ & p=6\Rightarrow x^2+y^2+6x+8y+9=0\\ & \Rightarrow \text{pusatnya adalah}(-{\displaystyle \frac{A}{2},-\frac{B}{2}})=(-3,-4)\\ & \text{dan berikut ilustrasi gambarnya}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Titik-titik berikut yang posisinya berada di luar }\\ & \text{lingkaran}{x}^2+y^2-2x+8y-32=0\text{adalah}....\\ & \text{A}.(0,0)\\ & \text{B}.(-6,-4)\\ & \text{C}.(-3,2)\\ & \text{D}.(3,1)\\ & \text{E}.(4,1)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \begin{array}{|cclc|}\hline \text{Opsi}& \text{Titik}& \text{Lingkaran}& \text{Keterangan}\\ \text{A}& (0,0)& 0^2+0^2-2.0+8.0-32=-32& \text{dalam}\\ \text{B}& (-6,-4)& (-6{)}^2+(-4{)}^2-2(-6)+8(-4)-32=0& \text{pada}\\ \text{C}& (-3,2)& (-3{)}^2+(2{)}^2-2(-3)+8(2)-32=3& \mathbf{\text{di luar}}\\ \text{D}& (3,1)& 3^2+1^2-2.3+8.1-32=-20& \text{dalam}\\ \text{E}& (4,1)& 4^2+1^2-2.4+8.1-32=-15& \text{dalam}\\ \hline\end{array}\\ & \text{Berikut ilustrasi gambarnya}\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Diketahui garis}x-2y=5\text{memotong lingkaran}\\ & x^2+y^2-4y+8y+10=0\text{di titik A dan B}.\\ & \text{Panjang ruas garis AB adalah}....\\ & \text{A}.4\sqrt{2}\\ & \text{B}.2\sqrt{5}\\ & \text{C}.\sqrt{10}\\ & \text{D}.5\\ & \text{E}.4\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \begin{array}{rl}& \text{Perhatikanlah bahwa garis}x-2y=5\\ & \text{memotong lingkaran}\\ & x^2+y^2-4x+8y+10=0,\\ & \text{maka garis}x=2y+5\text{disubstitusikan ke}\\ & \text{lingkaran tersebut, yaitu}:\\ & (2y+5{)}^2+y^2-4(2y+5)+8y+10=0\\ & 4y^2+20y+25+y^2-8y-20+8y+10=0\\ & 5y^2+20y+15=0\\ & y^2+4y+3=0\\ & (y+1)(y+3)=0\\ & y=-1\vee y=-3\\ & \text{untuk nilai}\\ & y=-3\Rightarrow x=2(-3)+5=-1,A(-1,-3)\\ & y=-1\Rightarrow x=2(-1)+5=3,B(3,-1)\\ & \text{maka},\text{AB}=\sqrt{(3-(-1){)}^2+(-1-(-3){)}^2}\\ & =\sqrt{4^2+2^2}\\ & =\sqrt{16+4}\\ & =\sqrt{20}\\ & =2\sqrt{5}\end{array}\\ & \text{Berikut ilustrasi gambarnya}\end{array}\end{array}$ .

$\begin{array}{l}\\ 10.& \text{Kekhususan persamaan lingkaran}\\ & x^2+y^2-6x-6y+6=0\text{adalah}....\\ & \text{A}.\text{menyinggung sumbu X}\\ & \text{B}.\text{menyinggung sumbu Y}\\ & \text{C}.\text{berpusat di}O(0,0)\\ & \text{D}.\text{titik pusatnya terletak pada}x-y=0\\ & \text{E}.\text{berjari-jari 3}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui persamaan lingkaran}\\ & x^2+y^2-6x-6y+6=0\\ & x^2-6x+9+y^2-6y+9+6=9+9\\ & (x-3{)}^2+(y-3{)}^2=18-6\\ & (x-3{)}^2+(y-3{)}^2=12\\ & (x-3{)}^2+(y-3{)}^2={\left(2\sqrt{3}\right)}^2\\ & \text{lingkaran ini}\{\begin{array}{ll}\text{Pusat}& =(3,3)\\ \text{Jari-jari}& =2\sqrt{3}\end{array}\\ & \begin{array}{|clc|}\hline \text{Opsi}& \text{Pernyataan}& \text{Keterangan}\\ \text{A}& \text{menyinggung sumbu X}& \text{tidak tepat}\\ \text{B}& \text{menyinggung sumbu Y}& \text{tidak tepat}\\ \text{C}& \text{berpusat di}O(0,0)& \text{tidak tepat}\\ \text{D}& \text{titik pusatnya terletak pada garis}x-y=0& \mathbf{\text{tepat}}\\ \text{E}& \text{berjari-jari 3}& \text{tidak tepat}\\ \hline\end{array}\\ & \text{Berikut ilustrasi gambarnya}\end{array}\end{array}$.