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Q.
If circle $(x-6)^{2}+y^{2}=r^{2}$ and parabola $y^{2}=4 x$ have maximum number of common chords, then least integral value of $r$ is _____
Conic Sections
Solution:
For maximum number of common chords, circle and parabola must intersect in $4$ distinct points.
Let's first find the value of $r$ when circle and parabola touch each other.
For that solving the given curves we have
or $(x-6)^{2}+4 x=r^{2} $
$x^{2}-8 x+36-r^{2}=0$
Curves touch if discriminant is 0 .
$D=64-4\left(36-r^{2}\right)=0 $ or $ r^{2}=20$
Hence least integral value of $r$ for which the curves intersect is $5 .$