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  4. Consider the parabola x2=4 y and circle x2+(y-5)2=r2(r>0). Given that the circle touches the parabola at the points P and Q. Let R be the point of intersection of tangents to parabola at P and Q and S be the centre of circle. The equation of the circle which passes through the vertex of the parabola x2=4 y and touches it at the point M(-4,4), is

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Q. Consider the parabola $x^2=4 y$ and circle $x^2+(y-5)^2=r^2(r>0)$. Given that the circle touches the parabola at the points $P$ and $Q$. Let $R$ be the point of intersection of tangents to parabola at $P$ and $Q$ and $S$ be the centre of circle.
The equation of the circle which passes through the vertex of the parabola $x^2=4 y$ and touches it at the point $M(-4,4)$, is

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Solution:

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The equation of tangent at $M(-4,4)$ to the parabola $x^2=4 y$, is
$x (-4)=2( y +4) \Rightarrow 2 x + y +4=0$
Let the required circle be $(x+4)^2+(y-4)^2+\lambda(2 x+y+4)=0$
As above circle passes through $(0,0)$, so $16+16+\lambda(4)=0 $
$\Rightarrow 4 \lambda=-32 \Rightarrow \lambda=-8$
Hence required equation of circle is $x^2+y^2-8 x-16 y=0$