Question

The circle $x^2+y^2=5$ meets the parabola $y^2=4x$ at $P$ and $Q$. Then the length $PQ$ is equal to

• A. $2$
• B. $2\sqrt{2}$
• C. $4$
• D. None of these

Answer 1

The correct option is C $4$

Solving circle $x^2+y^2=5$ and parabola $y^2=4x$, we have
$x^2+4x-5=0$
$\Rightarrow x=1$
or $x=-5$ (not possible)
$\Rightarrow y^2=4$ or $y=\pm 2$
$\Rightarrow$ Points of intersection are $P(1,2);Q(1,-2)$
Hence, $PQ=4$