$\begin{array}{ll}\\ 26.&\textrm{Diketahui lingkaran-lingkaran}\\ & x^{2}+y^{2}-2x+3y+k=0\: \: \textrm{dan}\: \\ &x^{2}+y^{2}+8x-6y-7=0\: \: \textrm{saling}\\ &\textrm{berpotongan ortogonal saat}\: \: k=\: ....\\ &\textrm{a}.\quad \color{red}-10\\ &\textrm{b}.\quad -3\\ &\textrm{c}.\quad 1\\ &\textrm{d}.\quad 5\\ &\textrm{e}.\quad 8\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan tabel berikut}\\ &\begin{array}{|l|l|l|}\hline \qquad\qquad\textrm{Lingakaran}&\qquad\textrm{Pusat/r}\\\hline L_{1}\equiv x^{2}+y^{2}-2x+3y+k=0&\begin{cases} P_{1} &=\left ( 1,-\displaystyle \frac{3}{2} \right ) \\ r_{1} & = \sqrt{\displaystyle \frac{13-4k}{4}} \end{cases}\\\hline \begin{aligned}L_{2}&\equiv x^{2}+y^{2}+8x-6y-7=0 \end{aligned}&\begin{cases} P_{2} &=\left ( -4,3 \right ) \\ r_{2} & = \sqrt{32} \end{cases}\\\hline \end{array} \\ &\textrm{Syarat dua lingkaran berpotongan ortogonal}\\ &\begin{aligned}&\left (P_{1}P_{2} \right )^{2}=r_{1}^{2}+r_{2}^{2}\\ &\Leftrightarrow \left ( 1+4 \right )^{2}+\left ( -\displaystyle \frac{3}{2}-3 \right )^{2}=\left ( \sqrt{\displaystyle \frac{13-4k}{4}} \right )^{2}+\sqrt{32}^{2}\\ &\Leftrightarrow \: 25+\displaystyle \frac{81}{4}=\displaystyle \frac{13-4k}{4}+32\\ &\Leftrightarrow \: 100+81=13-4k+128\\ &\Leftrightarrow \: k=-10 \end{aligned} \\ &\textbf{Sebagai ilustrasi perhatikan gambar berikut} \end{array}$.
$\begin{array}{ll}\\ 27.&\textrm{Persamaan lingkaran yang berpotongan}\\ &\textrm{lingkaran lain}\: \: x^{2}+y^{2}+2x+y-11=0\\ &\textrm{secara tegak lurus dan melalui}\: \: (4,3)\: \: \textrm{serta}\\ &\textrm{pusatnya pada}\: \: 9x+4y=37\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \color{red}x^{2}+y^{2}-10x+4y+3=0\\ &\textrm{b}.\quad x^{2}+y^{2}-8x+10y+6=0\\ &\textrm{c}.\quad x^{2}+y^{2}+4x-8y+7=0\\ &\textrm{d}.\quad x^{2}+y^{2}+6x+y+5=0\\ &\textrm{e}.\quad x^{2}+y^{2}+12x+6y+5=0\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan tabel berikut}\\ &\begin{array}{|l|l|l|}\hline \qquad\qquad\textrm{Lingakaran}&\qquad\textrm{Pusat/r}\\\hline L_{1}\equiv x^{2}+y^{2}+2x+y-11=0&\begin{cases} P_{1} &=\left ( -1,-\displaystyle \frac{1}{2} \right ) \\ r_{1} & = \sqrt{\displaystyle \frac{49}{4}}=\displaystyle \frac{7}{2} \end{cases}\\\hline \begin{aligned}L_{2}&\equiv (x-a)^{2}+(y-b)^{2}=r^{2} \end{aligned}&\begin{cases} P_{2} &=\left ( a,b \right ) \\ r_{2} & = r \end{cases}\\\hline \end{array}\\ &\textrm{Karena berpotongan tegak lurus, maka}\\ &\begin{aligned}&\left (P_{1}P_{2} \right )^{2}=r_{1}^{2}+r_{2}^{2}\\ &\Leftrightarrow \left ( -1-a \right )^{2}+\left ( -\displaystyle \frac{1}{2}-b \right )^{2}=\displaystyle \frac{49}{4}+r^{2}\\ &\Leftrightarrow a^{2}+2a+1+b^{2}+b+\displaystyle \frac{1}{4}=\displaystyle \frac{49}{4}+r^{2}\\ &\Leftrightarrow \color{blue}a^{2}+b^{2}+2a+b+\displaystyle \frac{5}{4}=\displaystyle \frac{49}{4}+r^{2}\\ &\Leftrightarrow a^{2}+b^{2}+2a+b-11=r^{2}\: .......(1)\\ \end{aligned} \\ &\textrm{Selanjutnya}\\ &\begin{aligned}&\textrm{Lingkaran}\: \: L_{2}\: \: \textrm{melalui titik}\: \: (4,3), \textrm{artinya}\\ &\textrm{bahwa}\: :\: (4-a)^{2}+(3-b)^{2}=r^{2}\\ &\Leftrightarrow a^{2}-8a+16+b^{2}-6b+9=r^{2}\\ &\Leftrightarrow a^{2}+b^{2}-8a-6b+25=r^{2}\: .......(2)\\ &\textrm{Pusat lingkaran}\: \: L_{2}\: \: \textrm{melalui garis}\: \: 9x+4y=37\\ &\textrm{artinya}:\: 9a+4b=37\: ...............(3)\\ \end{aligned}\\ &\begin{aligned}&\textrm{Dengan eliminasi}\: 1\: \&\: 2\: \: \textrm{dapat diperoleh}:\\ &\begin{array}{rll} a^{2}+b^{2}-8a-6b+25&=r^{2}&\\ a^{2}+b^{2}+2a+b-11&=r^{2}&-\\\hline -10a-7b+36&=0&\textrm{atau}\\ 10a+7b&=36&......(4) \end{array}\\ &\textrm{Dari persamaan}\: 3\: \&\: 4\: \: \textrm{dapat diperoleh}:\\ & \end{aligned}\\ &\begin{array}{rll} 10a+7b&=36&(\times 4)\\ 9a+4b&=37&(\times 7)\\\hline 40a+28b&=144&\\ 63a+28b&=259&\\\hline -23a\: \quad\quad&=-115&\\ a&=\displaystyle \frac{-115}{-23}&=5\\ 10(5)+7b&=36&\\ 7b&=-14\\ b&=-2 \end{array}\\ &\textrm{Adapun langkah berikutnya}\\ &\begin{aligned}&L_{2}\equiv (4-a)^{2}+(3-b)^{2}=r^{2}\\ &L_{2}\equiv (4-5)^{2}+(3+2)^{2}=r^{2}\\ &L_{2}\equiv r^{2}=25+1=26\\ &\textrm{Sehingga},\: L_{2}\equiv (x-5)^{2}+(y+2)^{2}=26\\ &\Leftrightarrow x^{2}+y^{2}-10x+4y+25+4-26=0\\ &\Leftrightarrow \color{red}x^{2}+y^{2}-10x+4y+3=0 \end{aligned}\\ &\textbf{Berikut ilustrasi gambarnya} \end{array}$.
$\begin{array}{ll}\\ 28.&\textrm{Diketahui lingkaran pertama berpusat di}\: \: (1,2)\\ &\textrm{dan menyinggung garis}\: \: 3x-4y+10=0.\\ &\textrm{Jika ada lingkaran kedua dengan pusat}\: \: (4,6)\\ &\textrm{dan menyinggung lingkaran yang pertama},\\ &\textrm{maka persamaan lingkaran yang kedua}\\ &\textrm{tersebut adalah}\: ....\\ &\textrm{a}.\quad x^{2}+y^{2}-8x-12y+48=0\\ &\textrm{b}.\quad x^{2}+y^{2}-8x-12y+43=0\\ &\textrm{c}.\quad \color{red}x^{2}+y^{2}-8x-12y+36=0\\ &\textrm{d}.\quad x^{2}+y^{2}-8x-12y+27=0\\ &\textrm{e}.\quad x^{2}+y^{2}-8x-12y+16=0\\\\ &\textbf{Jawab}:\\ &\textrm{Diketahui bahwa kedua lingkaran saling}\\ &\color{blue}\textrm{bersinggungan di luar},\: \color{black}\textrm{maka}\\ &\begin{aligned}r_{1}+r_{2}&=P_{1}P_{2}\\ &=\sqrt{(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2}}\\ &=\sqrt{(1-4)^{2}+(2-6)^{2}}\\ &=\sqrt{3^{2}+4^{2}}=\sqrt{5^{2}}=5 \end{aligned}\\ &\textrm{Selanjutnya}\\ &\begin{aligned}r_{\textrm{pertama}}&=\left |\displaystyle \frac{3(1)-4(2)+10}{\sqrt{3^{2}+4^{2}}} \right |\\ &=\left | \displaystyle \frac{3-8+10}{\sqrt{5^{2}}} \right |=\left | \displaystyle \frac{5}{5} \right |=\left | 1 \right |=1\\ \textrm{sehingga} &\\ r_{\textrm{kedua}}&=5-r_{\textrm{pertama}}=5-1=4 \end{aligned}\\ &\textrm{maka persamaan lingkaran keduanya adalah}:\\ &\begin{aligned}&(x-4)^{2}+(y-6)^{2}=4^{2}\\ &\Leftrightarrow x^{2}-8x+16+y^{2}-12y+36=16\\ &\Leftrightarrow \color{red}x^{2}+y^{2}-8x-12y+36=0 \end{aligned}\\ &\textbf{Berikut ilustrasi gambarnya} \end{array}$.
$\begin{array}{ll}\\ 29.&\textrm{Garis kuasa (tali busur sekutu)}\\ &\textrm{dari lingkaran}\\ &L_{1}\equiv x^{2}+y^{2}+6x-4y-12=0\\ &\textrm{dan}\: \: L_{2}\equiv x^{2}+y^{2}-12y=0\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad 3x+4y+9=0\\ &\textrm{b}.\quad 3x-4y-8=0\\ &\textrm{c}.\quad 3x-4y+7=0\\ &\textrm{d}.\quad 3x+4y-7=0\\ &\textrm{e}.\quad \color{red}3x+4y-6=0\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\\ &L_{1}\equiv x^{2}+y^{2}+6x-4y-12=0,\\ &\textrm{dan}\: \: L_{2}\equiv x^{2}+y^{2}-12y=0\\ &\textrm{Persamaan}\: \: \color{red}\textrm{garis kuasa}\: \color{black}\textrm{dari kedua}\\ &\textrm{lingkaran tersebut adalah}:\\ &\color{blue}L_{1}(x,y)- L_{2}(x,y)=0\\ &\Leftrightarrow x^{2}+y^{2}+6x-4y-12\\ &-(x^{2}+y^{2}-12y)=0\\ &\Leftrightarrow 6x+8y-12=0\\ &\Leftrightarrow \color{red}3x+4y-6=0 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 30.&\textrm{Jika dua lingkaran}\\ & x^{2}+y^{2}=9\: \: \textrm{dan}\\ &x^{2}+y^{2}-4y+2y+3=0\: \: \textrm{yang}\\ &\textrm{berpotongan di}\: \: (x_{1},y_{1})\: \: \textrm{dan}\: \: (x_{2},y_{2}),\\ &\textrm{maka nilai}\: \: 5(x_{1}+x_{2})\: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \color{red}24\\ &\textrm{b}.\quad 26\\ &\textrm{c}.\quad 28\\ &\textrm{d}.\quad 30\\ &\textrm{e}.\quad 32\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui}\\ &L_{1}\equiv x^{2}+y^{2}-9=0\: \: \textrm{dan}\\ &L_{2}\equiv x^{2}+y^{2}-4x+2y+3\\ &\textrm{Persamaan}\: \: \color{red}\textrm{garis kuasa}\: \color{black}\textrm{dari kedua}\\ &\textrm{lingkaran tersebut adalah}:\\ &\color{blue}L_{1}(x,y)- L_{2}(x,y)=0\\ &\Leftrightarrow x^{2}+y^{2}-9\\ &-(x^{2}+y^{2}-4y+2y+3)=0\\ &\Leftrightarrow 4x-2y-12=0\\ &\Leftrightarrow 2x-y-6=0\\ &\Leftrightarrow y=6-2x \end{aligned}\\ &\textrm{Selanjutnya}\\ &\begin{aligned}&x^{2}+y^{2}-9=0\\ &\Leftrightarrow x^{2}+(6-2x)^{2}-9=0\\ &\Leftrightarrow x^{2}+36-24x+4x^{2}-9=0\\ &\Leftrightarrow 5x^{2}-24x+27=0\\ &\Leftrightarrow x_{1,2}=\displaystyle \frac{24\pm \sqrt{576-540}}{10}\\ &\Leftrightarrow x_{1,2}=\displaystyle \frac{24\pm \sqrt{36}}{10}=\frac{24\pm 6}{10}\\ &\Leftrightarrow x_{1,2}=\displaystyle \frac{24\pm \sqrt{36}}{10}=\frac{24\pm 6}{10}\\ &\Leftrightarrow \quad x_{1}=3\: \: \textrm{atau}\: \: x_{2}=1,8\\ &\textrm{maka}\: \: 5(x_{1}+x_{2})=5\left ( 3+1,8 \right )=\color{red}24 \end{aligned} \end{array}$.
Tidak ada komentar:
Posting Komentar
Informasi