$\begin{array}{ll}\\ 31.&\textrm{Persamaan lingkaran yang melalui titik}\\ &(0,0)\: \: \textrm{dan titik potong kedua lingkaran}\\ &x^{2}+y^{2}-6x-8y-11=0\: \: \textrm{dan}\\ &x^{2}+y^{2}-4x-6y-22=0\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad x^{2}+y^{2}-12x+10y=0\\ &\textrm{b}.\quad x^{2}+y^{2}+8x-10y=0\\ &\textrm{c}.\quad x^{2}+y^{2}-8x+12y=0\\ &\textrm{d}.\quad \color{red}x^{2}+y^{2}-8x-10y=0\\ &\textrm{e}.\quad x^{2}+y^{2}+12x-8y=0\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}:\: \: L_{3}=L_{1}+p(L_{1}-L_{2})=0\\ &\textrm{dengan}\\ &\bullet \: L_{1}=x^{2}+y^{2}-6x-8y-11=0\\ &\bullet \: L_{2}=x^{2}+y^{2}-4x-6y-22=0\\ &\textrm{Untuk}\: \: L_{1}-L_{2}=-2x-2y+11=0\\ &\textrm{Karena}\: \: L_{3}\: \: \textrm{melalui}\: \: (0,0), \: \textrm{maka}\\ &\begin{aligned}L_{3}&=L_{1}+p(L_{1}-L_{2})=0\\ &=x^{2}+y^{2}-6x-8y-11 +p(-2x-2y+11)=0\\ &\Leftrightarrow 0^{2}+0^{2}-0-0-11+p(0+11)=0\\ &\Leftrightarrow p=\color{blue}1 \end{aligned}\\ &\textrm{Sehingga}\\ &L_{3}=x^{2}+y^{2}-6x-8y-11+(-2x-2y+11)=0\\ &\Leftrightarrow L_{3}=\color{red}x^{2}+y^{2}-8x-10y=0 \end{aligned} \end{array}$.
Berikut ilustrasi gambarnya
$\begin{array}{ll}\\ 32.&\textrm{Persamaan lingkaran yang melalui titik}\\ & (8,4)\: \: \textrm{dan titik potong lingkaran}\: x^{2}+y^{2}=16\\ &\textrm{dan}\: \: x^{2}+y^{2}-4x-4y=0\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad x^{2}+y^{2}-8x-8y-16=0\\ &\textrm{b}.\quad x^{2}+y^{2}-8x+8y+16=0\\ &\textrm{c}.\quad \color{red}x^{2}+y^{2}-8x-8y+16=0\\ &\textrm{d}.\quad x^{2}+y^{2}+8x+8y-16=0\\ &\textrm{e}.\quad x^{2}+y^{2}+8x+8y+16=0\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}:\: \: L_{3}=L_{1}+p(L_{1}-L_{2})=0\\ &\textrm{dengan}\\ &\bullet \: L_{1}=x^{2}+y^{2}-16=0\\ &\bullet \: L_{2}=x^{2}+y^{2}-4x-4y=0\\ &\textrm{Untuk}\: \: L_{1}-L_{2}=4x+4y-16=0\\ &\Leftrightarrow x+y=4\\ &\textrm{Karena}\: \: L_{3}\: \: \textrm{melalui}\: \: (8,4), \: \textrm{maka}\\ &\begin{aligned}L_{3}&=L_{1}+p(L_{1}-L_{2})=0\\ &=x^{2}+y^{2}-16+p(x+y-4)=0\\ &\Leftrightarrow 8^{2}+4^{2}-16+p(8+4-4)=0\\ &\Leftrightarrow -8p=\color{blue}64\color{black}\Leftrightarrow p=\color{blue}-8 \end{aligned}\\ &\textrm{Sehingga}\\ &L_{3}=x^{2}+y^{2}-16-8(x+y-4)=0\\ &\Leftrightarrow L_{3}=\color{red}x^{2}+y^{2}-8x-8y+16=0 \end{aligned}\\ &\textbf{Berikut ilustrasi gambarnyanya} \end{array}$.
$\begin{array}{ll}\\ 33.&\textrm{Persamaan lingkaran yang melalui titik}\\ & (7,-4)\: \: \textrm{dan titik potong kedua lingkaran}\\ &x^{2}+y^{2}-6x+8y-27=0\: \: \textrm{dan}\\ &x^{2}+y^{2}-26x+4y+121=0\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad x^{2}+y^{2}-36x-2y+121=0\\ &\textrm{b}.\quad x^{2}+y^{2}+24x-4y-222=0\\ &\textrm{c}.\quad 3x^{2}+3y^{2}-18x+2y-121=0\\ &\textrm{d}.\quad \color{red}x^{2}+y^{2}-36x+2y+195=0\\ &\textrm{e}.\quad x^{2}+y^{2}+24x+2y+195=0\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}:\: \: L_{3}=L_{1}+p(L_{1}-L_{2})=0\\ &\textrm{dengan}\\ &\bullet \: L_{1}=x^{2}+y^{2}-6x+8y-27=0\\ &\bullet \: L_{2}=x^{2}+y^{2}-26x+4y+121=0\\ &\textrm{Untuk}\: \: L_{1}-L_{2}=20x+4y-148=0\\ &\textrm{Karena}\: \: L_{3}\: \: \textrm{melalui}\: \: (7,-4), \: \textrm{maka}\\ &\begin{aligned}L_{3}&=L_{1}+p(L_{1}-L_{2})=0\\ &=x^{2}+y^{2}-6x+8y-27\\ &\qquad+p(20x+4y-148)=0\\ &\Leftrightarrow 7^{2}+(-4)^{2}-42-32-27\\ &\qquad+p(140-16-148)=0\\ &\Leftrightarrow -24p=\color{blue}36\color{black}\Leftrightarrow p=\color{blue}-\displaystyle \frac{3}{2} \end{aligned}\\ &\textrm{Sehingga}\\ &L_{3}=x^{2}+y^{2}-6x+8y-27\\ &\qquad-\displaystyle \frac{3}{2}(20x+4y-148)=0\\ &\Leftrightarrow L_{3}=\color{red}x^{2}+y^{2}-36x+2y+195=0 \end{aligned} \end{array}$.
Berikut ilustrasi gambarnya
$\begin{array}{ll}\\ 34.&\textrm{Persamaan lingkaran yang melalui perpotongan}\\&\textrm{lingkaran}\: \: x^{2}+y^{2}-12x+6y+20=0\: \: \textrm{dan}\\ &x^{2}+y^{2}-16x-14y+64=0\: \: \textrm{serta pusatnya}\\ &\textrm{terletak pada garis}\: \: 8x-3y-19=0\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \color{red}x^{2}+y^{2}-20x-34y+108=0\\ &\textrm{b}.\quad x^{2}+y^{2}-16x+12y+96=0\\ &\textrm{c}.\quad x^{2}+y^{2}-12x+20y+88=0\\ &\textrm{d}.\quad x^{2}+y^{2}+16x-24y+108=0\\ &\textrm{e}.\quad x^{2}+y^{2}+22x-34y+96=0\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa persamaan lingkaran}:\\ &\bullet \: L_{1}=x^{2}+y^{2}-12x+6y+20=0\\ &\bullet \: L_{2}=x^{2}+y^{2}-16x-14y+64=0\\ &\textrm{Persamaan tali busurnya (garis kuasa)}\\ &\textrm{adalah}:\\ &L_{1}(x,y)-L_{2}(x,y)\\ &=4x+20y-44=0\Leftrightarrow \color{blue}x=11-5y\\ &\textrm{Selanjutnya dengan substitusi }\\ &\begin{aligned}&x^{2}+y^{2}-12x+6y+20=0\\ &\Leftrightarrow (x-6)^{2}+(y+3)^{2}=25\\ &\Leftrightarrow (\color{blue}11-5y\color{black}-6)^{2}+(y+3)^{2}=25\\ &\Leftrightarrow (y-5y)^{2}+(y+3)^{2}=25\\ &\Leftrightarrow 26y^2-44y+9=0 \end{aligned}\\ &\textrm{Sehingga dengan}\: \: \color{red}\textrm{memodifikasi}\\ &\begin{aligned}&26y^2-44y+9=0\\ &\Leftrightarrow 25y^2-44y+y^2+9=0\\ &\quad\textrm{arahkan ke bentuk kuadrat sempurna}\\ &\Leftrightarrow 25y^2-10y+1+y^2-34y+8=0\\ &\Leftrightarrow 25y^2-10y+1+y^2-34y+17^{2}-17^{2}+8=0\\ &\Leftrightarrow (5y-1)^{2}+(y-17)^{2}-281=0\\ &\quad \textrm{ingat bahwa ada tali busur}\: \: \color{blue}5y=11-x\\ &\Leftrightarrow (\color{blue}11-x\color{black}-1)^{2}+(y-17)^{2}-281=0\\ &\Leftrightarrow (10-x)^{2}+(y-17)^{2}-281=0\\ &\Leftrightarrow x^{2}-20x+100+y^{2}-34y+289-281=0\\ &\Leftrightarrow \color{red}x^{2}+y^{2}-20x-34y+108=0 \end{aligned} \end{aligned}\\ &\textbf{Berikut ilustrasi gambarnya} \end{array}$
$\begin{array}{ll}\\ 35.&\textrm{Persamaan lingkaran dengan titik pusat}\\ &\textrm{pada garis}\: \: x+2y-3=0\: \: \textrm{dan melalui}\\ &\textrm{titik potong dua lingkaran}\\ &x^{2}+y^{2}-2x-4y+1=0\: \: \textrm{dan}\\ &x^{2}+y^{2}-4x-2y+4=0\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \color{red}x^{2}+y^{2}-6x+7=0\\ &\textrm{b}.\quad x^{2}+y^{2}-3y+4=0\\ &\textrm{c}.\quad x^{2}+y^{2}-2x-2y+1=0\\ &\textrm{d}.\quad x^{2}+y^{2}-2x-4y+4=0\\ &\textrm{e}.\quad x^{2}+y^{2}-3x-2y+7=0\\\\ &\textbf{Jawab}:\\ &\color{blue}\textbf{Alternatif 1}\\ &\textrm{Gunakan cara pembahasan sebagaimana pada}\\ &\textrm{nomor-nomor sebelumnya}\\ &\color{blue}\textbf{Alternatif 2}\\ &\begin{aligned}&\textrm{Diketahui}\\ &L_{1}\equiv x^{2}+y^{2}-2x-4y+1=0,\: \: \textrm{dan}\\ &L_{2}\equiv x^{2}+y^{2}-4x-2y+4=0\\ &\textrm{Persamaan}\: \: \color{red}\textrm{tali busur}\: \color{black}\textrm{dari kedua}\\ &\textrm{lingkaran tersebut adalah}:\\ &\color{blue}L_{1}(x,y)- L_{2}(x,y)=0\\ &\Leftrightarrow x^{2}+y^{2}-2x-4y+1\\ &-(x^{2}+y^{2}-4x-2y+4)=0\\ &\Leftrightarrow 2x-2y-3=0\\ &\textrm{Selanjutnya perlu ditentukan juga}\\&\textrm{Persamaan}\: \: \color{red}\textrm{berkas lingkaran}\: \color{black}\textrm{melalui}\\ &\textrm{titik-titik potong kedua lingkaran}\\ &\textrm{di atas adalah}:\\ &L_{1}+\lambda L_{2}=0\\ &x^{2}+y^{2}-2x-4y+1\\ &\qquad+\lambda \left ( x^{2}+y^{2}-4x-2y+4 \right )=0\\ &\Leftrightarrow (1+\lambda )x^{2}+(1+\lambda )y^{2}-(2+4\lambda )x\\ &\qquad -(4+2\lambda )y+1+4\lambda =0\\ &\textrm{Saat}\: \: \lambda =-1,\: \textrm{maka persamaan berkas}\\ &\textrm{lingkarannya adalah}:\: 2x-2y-3=0\\ &\textrm{Hal ini hasilnya sama persis saat kita}\\ &\textrm{menentukan persamaan}\: \color{red}\textrm{tali busur}\: \color{black}\textrm{di atas}\\ &\textrm{Selanjutnya kita ambil}\\ &L_{2}-(L_{1}+\lambda L_{2})=0\\ &\Leftrightarrow x^{2}+y^{2}-4x-2y+4-(2x-2y-3)=0\\ &\Leftrightarrow \color{red}x^{2}+y^{2}-6x+7=0 \end{aligned} \end{array}$.
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Lingkaran baru yang berpusat di (3,0)
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