Question

A circle with centre at $(15,-3)$ is tangent to $y=\frac{x^2}{3}$ at a point in the first quadrant. The radius of the circle is equal to $a\sqrt{5}$ where $a$ is

Answer 1

Let the point of tangency be $(x_1,\frac{x_1^2}{3})$
Slope of the tangent, $m_t=y^{\prime }=\frac{2x_1}{3}$
$\therefore$ Slope of the normal, $m_n=-\frac{3}{2x_1}$
$\Rightarrow \frac{\frac{x_1^2}{3}+3}{x_1-15}=-\frac{3}{2x_1}$
$\Rightarrow 2x_1(x_1^2+9)=9(15-x_1)$
$\Rightarrow 2x_1^3+18x_1=135-9x_1$
$\Rightarrow 2x_1^3+27x_1-135=0$
$\Rightarrow (x_1-3)(2x_1^2+6x_1+45)=0$
$\Rightarrow x_1=3$ or $2x_1^2+6x_1+45=0$

$2x_1^2+6x_1+45=0(D<0)$
So, $(x_1,y_1)=(3,3)$
The radius of the circle will be
$r=\sqrt{(15-3{)}^2+(-3-3{)}^2}\Rightarrow r=6\sqrt{5}=a\sqrt{5}$

Hence, the value of $a$ is $6$