$\begin{array}{ll}\\ 6.&\textrm{Diketahui lingkaran}\: \: x^{2}+y^{2}+4x+ky-12=0\\ &\textrm{melalui titik}\: \: (-2,8)\: \: \textrm{maka jari-jari lingkaran}\\ &\textrm{tersebut adalah}....\\ &\textrm{A}.\quad 1\\ &\textrm{B}.\quad \color{red}5\\ &\textrm{C}.\quad 6\\ &\textrm{D}.\quad 12\\ &\textrm{E}.\quad 25\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui ingkaran berpusat di}\: \left ( -2,-\displaystyle \frac{1}{2}k \right ),\\ &\textrm{yaitu}:\\ &x^{2}+y^{2}+4x+ky-12=0\\ & \textrm{melalui}\: \: (-2,8)\: \: \textrm{berarti }\\ &(-2)^{2}+8^{2}+4(-2)+k.8-12=0\\ &4+64-8-12+8k=0\\ &48+8k=0\\ &k=\color{blue}-6\\ &\textrm{Sehingga}\: \: r=\sqrt{\displaystyle \frac{4^{2}}{4}+\frac{(-6)^{2}}{4}-(-12)}\\ &\qquad\qquad \: \: \: =\sqrt{\displaystyle 4+9+12}=\sqrt{25}=\color{red}5\\ \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 7.&\textrm{Persmaan lingkaran}\: \: x^{2}+y^{2}+px+8y+9=0\\ &\textrm{menyinggung sumbu X. Pusat lingkaran tersebut }\\ &\textrm{adalah}\: ....\\ &\textrm{A}.\quad (6,-4)\\ &\textrm{B}.\quad (6,6)\\ &\textrm{C}.\quad \color{red}(3,-4)\\ &\textrm{D}.\quad (-6,-4)\\ &\textrm{E}.\quad (3,4)\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textbf{Lingkaran}\: \: x^{2}+y^{2}+px+8y+9=0\\ &\textrm{maka,}\\ &x^{2}+px+y^{2}+8y+9=0\\ &\left ( x+\displaystyle \frac{1}{2}p \right )^{2}-\displaystyle \frac{1}{4}p^{2}+(y+4)^{2}-16+9=0\\ &\Leftrightarrow \left ( x+\displaystyle \frac{1}{2}p \right )^{2}+(y+4)^{2}=7+\displaystyle \frac{1}{4}p^{2}\\ &\textrm{karena menyinggung sumbu-X,}\: \: \: \: R=b=4,\\ & \textrm{sehingga}\\ &7+\displaystyle \frac{1}{4}p^{2}=4^{2}\\ &\Leftrightarrow \displaystyle \frac{1}{4}p^{2}=16-7=9\Leftrightarrow p^{2}=36\Leftrightarrow p=\color{blue}\pm 6\\ &p=-6\: \Rightarrow \: x^{2}+y^{2}-6x+8y+9=0\\ &\quad\Rightarrow \textrm{pusatnya adalah}\: \: \left ( -\displaystyle \frac{A}{2},-\frac{B}{2} \right )=\color{red}(3,-4)\\ &p=6\: \: \: \, \: \Rightarrow \: x^{2}+y^{2}+6x+8y+9=0\\ &\quad\Rightarrow \textrm{pusatnya adalah}\: \: \left ( -\displaystyle \frac{A}{2},-\frac{B}{2} \right )=\color{red}(-3,-4)\\ &\color{purple}\textrm{dan berikut ilustrasi gambarnya} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 8.&\textrm{Titik-titik berikut yang posisinya berada di luar }\\ &\textrm{lingkaran}\: \: x^{2}+y^{2}-2x+8y-32=0\: \: \textrm{adalah}.... \\ &\textrm{A}.\quad (0,0)\\ &\textrm{B}.\quad (-6,-4)\\ &\textrm{C}.\quad \color{red}(-3,2)\\ &\textrm{D}.\quad (3,1)\\ &\textrm{E}.\quad (4,1)\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\begin{array}{|c|c|l|c|}\hline \color{blue}\textrm{Opsi}&\color{blue}\textrm{Titik}&\qquad\qquad\quad\color{blue}\textrm{Lingkaran}&\color{blue}\textrm{Keterangan}\\\hline \textrm{A}&(0,0)&0^{2}+0^{2}-2.0+8.0-32=-32&\textrm{dalam}\\\hline \textrm{B}&(-6,-4)&(-6)^{2}+(-4)^{2}-2(-6)+8(-4)-32=0&\textrm{pada}\\\hline \color{red}\textrm{C}&(-3,2)&(-3)^{2}+(2)^{2}-2(-3)+8(2)-32=3&\textbf{di luar}\\\hline \textrm{D}&(3,1)&3^{2}+1^{2}-2.3+8.1-32=-20&\textrm{dalam}\\\hline \textrm{E}&(4,1)&4^{2}+1^{2}-2.4+8.1-32=-15&\textrm{dalam}\\\hline \end{array} \\ &\color{purple}\textrm{Berikut ilustrasi gambarnya} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 9.&\textrm{Diketahui garis}\: \: x-2y=5\: \: \textrm{memotong lingkaran}\\ &x^{2}+y^{2}-4y+8y+10=0\: \: \textrm{di titik A dan B}.\\ &\textrm{Panjang ruas garis AB adalah}....\\ &\textrm{A}.\quad 4\sqrt{2}\\ &\textrm{B}.\quad \color{red}2\sqrt{5}\\&\textrm{C}.\quad \sqrt{10}\\ &\textrm{D}.\quad 5\\ &\textrm{E}.\quad 4\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\begin{aligned}&\textrm{Perhatikanlah bahwa garis}\: \: \color{blue}x-2y=5\\&\textrm{memotong lingkaran}\\ &x^{2}+y^{2}-4x+8y+10=0,\\ &\textrm{maka garis}\: \: \color{blue}x=2y+5\: \: \color{black}\textrm{disubstitusikan ke}\\ &\textrm{lingkaran tersebut, yaitu}:\\ &(\color{blue}2y+5\color{black})^{2}+y^{2}-4(\color{blue}2y+5\color{black})+8y+10=0\\ &4y^{2}+20y+25+y^{2}-8y-20+8y+10=0\\ &5y^{2}+20y+15=0\\ &y^{2}+4y+3=0\\ &(y+1)(y+3)=0\\ &y=-1\: \: \vee \: \: y=-3\\ &\textrm{untuk nilai}\\ & y=-3\Rightarrow x=2(-3)+5=-1,\quad A(-1,-3)\\ &y=-1\Rightarrow x=2(-1)+5=3,\qquad B(3,-1)\\ &\textrm{maka},\qquad \textrm{AB}=\sqrt{(3-(-1))^{2}+(-1-(-3))^{2}}\\ &=\sqrt{4^{2}+2^{2}}\\ &=\sqrt{16+4}\\ &=\sqrt{20}\\ &=\color{red}2\sqrt{5} \end{aligned}\\ &\color{purple}\textrm{Berikut ilustrasi gambarnya} \end{aligned} \end{array}$ .
$\begin{array}{ll}\\ 10.&\textrm{Kekhususan persamaan lingkaran}\\ &x^{2}+y^{2}-6x-6y+6=0\: \: \textrm{adalah}....\\ &\textrm{A}.\quad \textrm{menyinggung sumbu X}\\ &\textrm{B}.\quad \textrm{menyinggung sumbu Y}\\ &\textrm{C}.\quad \textrm{berpusat di}\: \: O(0,0)\\ &\textrm{D}.\quad \color{red}\textrm{titik pusatnya terletak pada}\: \: x-y=0\\ &\textrm{E}.\quad \textrm{berjari-jari 3}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui persamaan lingkaran}\\ &x^{2}+y^{2}-6x-6y+6=0\\ &x^{2}-6x+9+y^{2}-6y+9+6=9+9\\ &(x-3)^{2}+(y-3)^{2}=18-6\\ &(x-3)^{2}+(y-3)^{2}=12\\ &(x-3)^{2}+(y-3)^{2}=\left ( 2\sqrt{3} \right )^{2}\\ &\textrm{lingkaran ini}\begin{cases} \textrm{Pusat} &=\color{blue}(3,3) \\ \textrm{Jari-jari} &=\color{blue}2\sqrt{3} \end{cases}\\ &\begin{array}{|c|l|c|}\hline \textrm{Opsi}&\qquad\qquad\qquad\textrm{Pernyataan}&\textrm{Keterangan}\\\hline \textrm{A}&\textrm{menyinggung sumbu X}&\textrm{tidak tepat}\\\hline \textrm{B}&\textrm{menyinggung sumbu Y}&\textrm{tidak tepat}\\\hline \textrm{C}&\textrm{berpusat di}\: \: O(0,0)&\textrm{tidak tepat}\\\hline \color{red}\textrm{D}&\color{red}\textrm{titik pusatnya terletak pada garis}\: \: x-y=0&\textbf{tepat}\\\hline \textrm{E}&\textrm{berjari-jari 3}&\textrm{tidak tepat}\\\hline \end{array} \\ &\textrm{Berikut ilustrasi gambarnya} \end{aligned} \end{array}$.
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