Question

Radius of the circle that passes through origin and touches the parabola $y^2=4ax$ at the point $(a,2a)$ is

• A. $\frac{5}{\sqrt{2}}a$
• B. $2\sqrt{2}a$
• C. $\sqrt{\frac{5}{2}}a$
• D. $\frac{3}{\sqrt{2}}a$

Answer 1

The correct option is A $\frac{5}{\sqrt{2}}a$

Given equation of parabola is
$y^2=4ax$
$\Rightarrow \frac{dy}{dx}=\frac{2a}{y}$
Slope of tangent at $(a,2a)$ = Slope of parabola at $(a,2a)=1$
Equation of tangent to parabola at $(a,2a)$ is
$y-2a=(x-a)$
or,$y-x-a=0$
Equation of circle touching the parabola at $(a,2a)$ is
$(x-a{)}^2+(y-2a{)}^2+\lambda (y-x-a)=0$
Since this circle passes through origin
$a^2+4a^2-a\lambda =0$
$\Rightarrow \lambda =5a$
So, the equation of circle is
$(x-a{)}^2+(y-2a{)}^2+5a(y-x-a)=0$
$\Rightarrow x^2+y^2-7ax+ay=0$
Comparing with general equation of circle
$x^2+y^2+2gx+2fy+c=0$
So, $g=-\frac{7a}{2},f=\frac{a}{2},c=0$
Radius $=\sqrt{\frac{49a^2}{4}+\frac{a^2}{4}}=\frac{5a}{\sqrt{2}}$