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Q.
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
Application of Derivatives
Solution:
$y=\left.\frac{x^2}{3} \Rightarrow \frac{d y}{d x}\right|_P=\frac{2 x_1}{3}$
$\therefore m _{ N }=-\frac{3}{2 x _1}=\frac{ y _1+3}{ x _1-15} \Rightarrow 2 x _1 y _1+6 x _1=-3 x _1+45$
Put $y _1=\frac{ x _1^2}{3}$, we get $2 x _1 \cdot \frac{ x _1^2}{3}+9 x _1-45=0 ; 2 x _1^3+27 x _1-135=0$
$2 x_1^2\left(x_1-3\right)+6 x_1\left(x_1-3\right)+45\left(x_1-3\right)=0 $
$\left(x_1-3\right)\left(2 x_1^2+6 x_1+45\right)=0 \Rightarrow x_1=3$
