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. $5\sqrt{6}$
• B. $8\sqrt{3}$
• C. $9\sqrt{2}$
• D. $6\sqrt{5}$

Answer 1

Answer: (D)

$y={\frac{x^2}{3}\Rightarrow \frac{dy}{dx}|}_P=\frac{2x_1}{3}$
$\therefore m_N=-\frac{3}{2x_1}=\frac{y_1+3}{x_1-15}\Rightarrow 2x_1{y}_1+6x_1=-3x_1+45$
Put $y_1=\frac{x_1^2}{3}$, we get $2x_1\cdot \frac{x_1^2}{3}+9x_1-45=0;2x_1^3+27x_1-135=0$
$2x_1^2\left(x_1-3\right)+6x_1\left(x_1-3\right)+45\left(x_1-3\right)=0$
$\left(x_1-3\right)\left(2x_1^2+6x_1+45\right)=0\Rightarrow x_1=3$