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Question Number 144201 by bramlexs22 last updated on 23/Jun/21

$$\mathrm{Find}\mathrm{the}\mathrm{equations}\mathrm{of}\mathrm{the}\mathrm{circles}$$$$\mathrm{passing}\mathrm{through}(-4,3)\mathrm{and}\mathrm{touching}$$$$\mathrm{the}\mathrm{lines}\mathrm{x}+\mathrm{y}=2\mathrm{and}\mathrm{x}-\mathrm{y}=2$$

Answered by benjo_mathlover last updated on 23/Jun/21

$$\mathrm{let}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{circle}\mathrm{be}$$$${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0...\left(\mathrm{i}\right)$$$$\mathrm{it}\mathrm{passes}\mathrm{through}(-4,3)\mathrm{therefore}$$$$25-8\mathrm{g}+6\mathrm{f}+\mathrm{c}=0...\left(\mathrm{ii}\right)$$$$\mathrm{since}\mathrm{circle}\mathrm{touches}\mathrm{the}\mathrm{lines}$$$$\{\begin{array}{l}\mathrm{x}+\mathrm{y}-2=0\\ \mathrm{x}-\mathrm{y}-2=0\end{array}\Rightarrow \mathrm{then}$$$$\mid \frac{-\mathrm{g}-\mathrm{f}-2}{\sqrt{2}}\mid =\mid \frac{-\mathrm{g}+\mathrm{f}-2}{\sqrt{2}}\mid =\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}...\left(\mathrm{iii}\right)$$$$\mathrm{Now}\mid \frac{-\mathrm{g}-\mathrm{f}-2}{\sqrt{2}}\mid =\mid \frac{-\mathrm{g}+\mathrm{f}-2}{\sqrt{2}}\mid$$$$\mathrm{we}\mathrm{get}\mathrm{f}=0\mathrm{or}\mathrm{g}=-2$$$$\mathrm{case}\left(1\right)\mathrm{for}\mathrm{f}=0$$$$\Rightarrow \mid \frac{-\mathrm{g}-2}{\sqrt{2}}\mid =\sqrt{{\mathrm{g}}^2-\mathrm{c}}$$$$\Rightarrow {(\mathrm{g}+2)}^2=2({\mathrm{g}}^2-\mathrm{c})$$$$\Rightarrow {\mathrm{g}}^2-4\mathrm{g}-2\mathrm{c}-4=0...\left(\mathrm{iv}\right)$$$$\mathrm{putting}\mathrm{f}=0\mathrm{in}\left(\mathrm{ii}\right)\mathrm{gives}25-8\mathrm{g}+\mathrm{c}=0...\left(\mathrm{v}\right)$$$$\mathrm{eliminaring}\mathrm{c}\mathrm{between}\left(\mathrm{iv}\right)\mathrm{\&}\left(\mathrm{v}\right)$$$${\mathrm{g}}^2-20\mathrm{g}+46=0;\mathrm{g}=10\pm 3\sqrt{6}$$$$\mathrm{substituting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{g}\mathrm{in}\left(\mathrm{v}\right)$$$$\mathrm{we}\mathrm{find}\mathrm{c}=55\pm 24\sqrt{6}$$$$\mathrm{so}\mathrm{the}\mathrm{eq}\mathrm{of}\mathrm{circle}\mathrm{is}$$$${\mathrm{x}}^2+{\mathrm{y}}^2\pm 2(10\pm 3\sqrt{6})\mathrm{x}+(55\pm 24\sqrt{6})=0$$$$$$

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