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Question Number 140325 by liberty last updated on 06/May/21

$$$$ What is the equation of the circle, if the circle is tangential to the line 3x+y+2=0 at (-1,1) and it passes through the point (3,5)?\n

Answered by benjo_mathlover last updated on 06/May/21

$$\left(1\right)\mathrm{let}(\mathrm{a},\mathrm{b})\mathrm{is}\mathrm{the}\mathrm{center}\mathrm{point}\mathrm{of}$$ $$\mathrm{circle}$$ $$\left(2\right)\left(\frac{\mathrm{b}-1}{\mathrm{a}+1}\right).(-3)=-1$$ $$\Rightarrow 3\mathrm{b}-3=\mathrm{a}+1;\mathrm{a}=3\mathrm{b}-4$$ $$\left(3\right)\frac{\mid 3\mathrm{a}+\mathrm{b}+2\mid }{\sqrt{10}}=\sqrt{{(\mathrm{a}-3)}^2+{(\mathrm{b}-5)}^2}$$ $$\Rightarrow \mid 10\mathrm{b}-10\mid =\sqrt{10}\sqrt{{(3\mathrm{b}-7)}^2+{(\mathrm{b}-5)}^2}$$ $$\Rightarrow 10({\mathrm{b}}^2-2\mathrm{b}+1)={10\mathrm{b}}^2-10\mathrm{b}-42\mathrm{b}+74$$ $$\Rightarrow -20\mathrm{b}+10=-52\mathrm{b}+74$$ $$\Rightarrow 32\mathrm{b}=64\to \{\begin{array}{l}\mathrm{b}=2\\ \mathrm{a}=2\end{array}$$ $$\left(4\right)\mathrm{radius}=\sqrt{{(2-3)}^2+{(2-5)}^2}=\sqrt{10}$$ $$\left(5\right)\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{circle}$$ $$\text{Extra close brace or missing open brace}$$

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