Question

Radius of the largest circle which passes through the focus of the parabola $y^2=4x$ and contained in it is

• A. $8$
• B. $4$
• C. $2$
• D. $5$

Answer 1

Answer: (B)

$4$

Let $(h,0)$ be the circle’s center.

Here,

$h=1+r$

Equation of the circle is,

${(x-h)}^2+y^2=r^2$

${(x-1-r)}^2+y^2=r^2$

$x^2+y^2-2x-2xr+1=0$

But, $y^2=4x$ (given). Thus, we get

$x^2+2x-2xr+2r+1=0$

$x^2+x(2-2r)+2r+1=0$

The above quadratic equation will have equal roots, because the circle intersects the parabola. So,

${(2-2r)}^2=4(2r+1)$

$4+4r^2-8r=8r+4$

$4r^2-16r=0$

$r^2-4r=0$

$r(r-4)=0$

Radius cannot be zero. So,

$r=4$

Hence, this is the required result.