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  4. The circumference of the circle x2+y2-2 x+8 y-q=0 is bisected by the circle x2+y2+4 x+12 y+p=0, then p+q is equal to:

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Q. The circumference of the circle $x^2+y^2-2 x+8 y-q=0$ is bisected by the circle $x^2+y^2+4 x+12 y+p=0$, then $p+q$ is equal to:

Conic Sections

Solution:

Common chord of given circle
$6 x+4 y+(p+q)=0$
image
This is diameter of $x^2+y^2-2 x+8 y-q=0$ centre $(1,-4)$
$6-16+(p+q)=0 \Rightarrow p+q=10$