Equation of the chord of the circle $x^2+y^2-4 x-10 y+25=0$ having mid-point $(1,2)$ is

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Aicharya · Accepted Answer

$\begin{aligned} & x^2+y^2-4 x-10 y+25=0 \Rightarrow 	ext { centre }=(2,5) 	ext { and } \ & 	ext { radius }=\sqrt{4+25-25}=2 \end{aligned}$ Refer figure Let $\mathrm{M}(1,2)$ be the midpoint of chord Slope of $\mathrm{CM}=\frac{2-5}{1-2}=3$ $	herefore$ Slope of $\mathrm{AB}=\frac{-1}{3}$ Equation of $\mathrm{AB}$ is $(\mathrm{y}-2)=\frac{-1}{3}(\mathrm{x}-1)$ i.e. $\mathrm{x}+3 \mathrm{y}=7$

Equation of the chord of the circle $x^2+y^2-4 x-10 y+25=0$ having mid-point $(1,2)$ is

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