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Question Number 135000 by bobhans last updated on 09/Mar/21

$$$$ Find the equation of the circle through the points of intersection of x^2+y^2−1=0,x^2+y^2−2x−4y+1=0 and touching the line x+2y=0?\n

Answered by EDWIN88 last updated on 09/Mar/21

$$\mathrm{Let}\mathrm{C}\mathrm{is}\mathrm{equation}\mathrm{of}\mathrm{circle},\mathrm{so}$$ $$\mathrm{C}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2-1+\lambda (-2\mathrm{x}-4\mathrm{y}+2)=0$$ $$\equiv {\mathrm{x}}^2+{\mathrm{y}}^2-2\lambda \mathrm{x}-4\lambda \mathrm{y}+2\lambda -1=0$$ $$\mathrm{witb}\mathrm{center}\mathrm{point}\mathrm{at}(\lambda ,2\lambda )\mathrm{and}\mathrm{radius}$$ $$\mathrm{r}=\sqrt{{\lambda }^2+4{\lambda }^2+1-2\lambda }=\frac{\mid \lambda +4\lambda \mid }{\sqrt{5}}$$ $$\Rightarrow \sqrt{5(5{\lambda }^2+1-2\lambda )}=\mid 5\lambda \mid$$ $$\Rightarrow 25{\lambda }^2+5-10\lambda =25{\lambda }^2;\lambda =\frac{1}{2}$$ $$\mathrm{s}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{g}\mathrm{e}\mathrm{t}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\equiv {\mathrm{x}}^2+{\mathrm{y}}^2-\mathrm{x}-2\mathrm{y}=0$$

Commented bybobhans last updated on 09/Mar/21

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