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

Answer 1

Step 1: Find the point of intersection of given circle and parabola

Solving circle$x^2+y^2=5$ and parabola $y^2=4x$, we have
$\begin{array}{rcl}{x}^2+4x-5& =& 0\\ x^2+5x-x-5& =& 0\\ x\left(x+5\right)-1\left(x+5\right)& =& 0\\ \left(x+5\right)\left(x-1\right)& =& 0\\ x& =& -5,x=1\end{array}$

When $x=-5$

$y^2=-20$, which is not possible for any real number.

When $x=1$
$y^2=4$ or $y=\pm 2$
So, Points of intersection are $P\left(1,2\right),Q\left(1,-2\right)$

Step 2: Use distance formula

The distance between any two points $A\left(x_1,y_1\right)$ and $B\left(x_2,y_2\right)$ is given by

$AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Therefore, the distance between $P\left(1,2\right),$ and $Q\left(1,-2\right)$ is

$\begin{array}{rcl}PQ& =& \sqrt{{\left(1-1\right)}^2+{\left(-2-2\right)}^2}\\ & =& \sqrt{4^2}\\ & =& 4\mathrm{unit}\end{array}$
Hence, $PQ$ is equal to $4\mathrm{unit}$.