If the mid point of the chord intercepted by the circle $x^2+y^2-8 x+10 y+5=0$ on the line $2 x+y+2=0$ is $(\mathrm{h}, \mathrm{k})$, then $\mathrm{k}+4 \mathrm{~h}=$

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Aicharya · Accepted Answer

Given the equation of circle $\begin{aligned} & x^2+y^2-8 x+10 y+5=0 \\ & \text { Centre }=(4,-5) \end{aligned}$

Now, equation of chord is $2 \mathrm{x}+\mathrm{y}+2=0$ since $(\mathrm{h}, \mathrm{k})$ is mid point of chord $\Rightarrow 2 \mathrm{~h}+\mathrm{k}+2=0$ ... (i) Now, slope of $\mathrm{AB} \times$ Slope of $\mathrm{PC}=-1$ $\begin{aligned} & \Rightarrow-2 \times\left(\frac{\mathrm{k}+5}{\mathrm{~h}-4}\right)=-1 \\ & \Rightarrow 2 \mathrm{k}+10=\mathrm{h}-4 \Rightarrow \mathrm{h}-2 \mathrm{k}=14 .... (ii) \end{aligned}$ On solving (i) \& (ii), we get $\mathrm{h}=2, \mathrm{k}=-6$ Now $k+4 h=-6+8=2$

If the mid point of the chord intercepted by the circle $x^2+y^2-8 x+10 y+5=0$ on the line $2 x+y+2=0$ is…

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