Question

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

• A. ${(x-2)}^2+y^2=9$
• B. ${(x-2)}^2+y^2=3$
• C. ${(x-2)}^2+y^2=2$
• D. ${(x-2)}^2+y^2=1$

Answer 1

The correct option is A ${(x-2)}^2+y^2=9$

$y^2=8x$

This eqn. is of the form

$y^2=4ax\Rightarrow 4a=8$

$a=2$

$\therefore$ Its focus $S$ is $(2,0)$.

Clearly the points $P$ and $Q$

are points of intersection of

$y^2=8x$ and $x^2+y^2-2x-4y=0$

$\downarrow$ $\downarrow$

$\left(1\right)$ $\left(2\right)$

Now we solve (1) and (2) to

find $p$ and $Q$