Question

The line $y=x+1$ meets the ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$ at two points $P$ and $Q$. If $r$ is the radius of the circle with $PQ$ as diameter then $(3r{)}^2$ is equal to

• A. 20
• B. 12
• C. 11
• D. 8

Answer 1

Answer: (A)

Ellipse $x^2+2y^2=4$
Line $y=x+1$
Point of intersection
$x^2+2(x+1{)}^2=4$
$3x^2+4x-2=0$
$\left|x_1-x_2\right|=\frac{\sqrt{40}}{3}$
$AB=2r=\left|x_1-x_2\right|\sqrt{1+m^2}$
$m$ is slope of given line
$AB=\frac{\sqrt{40}}{3}\sqrt{1+1}$
$2r=\frac{\sqrt{80}}{3}$
$\Rightarrow r=\frac{\sqrt{80}}{6}$
$(3r{)}^2={\left(3\times \frac{\sqrt{80}}{6}\right)}^2$
$=\frac{80}{4}=20$