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  4. From the focus of the parabola y 2=8 x as centre, a circle is described so that a common chord of the curves is equidistant from the vertex and focus of the parabola. The equation of the circle is -

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Q. From the focus of the parabola $y ^2=8 x$ as centre, a circle is described so that a common chord of the curves is equidistant from the vertex and focus of the parabola. The equation of the circle is -

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Solution:

Focus of parabola $y^2=8 x$ is $(2,0)$. Equation of circle with centre $(2,0)$ is $(x-2)^2+y^2=r^2$
$AB$ is common chord
$Q$ is mid point i.e. $(1,0)$
image
$AQ ^2= y ^2$ where $y ^2=8 1=8$
$\therefore r ^2= AQ ^2+ QS ^2=8+1=9$
so circle is $(x-2)^2+y^2=9$