Question

A circle with radius $r$ touches the parabola $y^2=4x$ at two points such that area formed by common tangent and common chord is $32$ sqaure units. the value of $r$ is

• A. $\sqrt{5}$ units
• B. $2\sqrt{5}$units
• C. $3\sqrt{5}$units
• D. $4\sqrt{5}$units

Answer 1

The correct option is B $2\sqrt{5}$units


For circle to touch the parabola at two distinct points,the circle should be symmetric to the axis of the parabola as given in the figure

For given parabola $y^2=4x$
$\Rightarrow a=1$

Let $B(t^2,2t),C(t^2,-2t)$.So, $D\equiv (t^2,0)$
Tangents at $B,C$ intersect at $A\equiv (-t^2,0)$

$\therefore$ area of required$△ABC=2\times \frac{1}{2}\times DB\times AD$
$\Rightarrow 32=\left(2t\right)\times \left(2t^2\right)$
$\Rightarrow t=2$

Now, normal at $B$ will pass through the centre of circle

equation of Normal is
$y+xt=2t{+}^3$
$\Rightarrow y+2x=4+8$
$\Rightarrow y+2x=12$

for $y=0$
$x=6$
Hence, centre of circle is $(6,0)$

Radius of this circle is $OB=\sqrt{(6-4{)}^2+(0-4{)}^2}=\sqrt{4+16}=2\sqrt{5}$ units