Question

Consider, $S : \frac{ x ^2}{9}-\frac{ y ^2}{16}=1$ and $C _1, C _2$ be two another curves defined as
$C _1$ : Locus of centre of a circle which cuts the curve $S =0$ orthogonally at $P \left(5, \frac{16}{3}\right)$.
$C _2$ : Locus of centre of a circle which touches the curve $S =0$ at $P \left(5, \frac{16}{3}\right)$.
Equation of a circle which touches $C _1$ and centred at the origin, is

• A. $x^2+y^2=9$
• B. $x^2+y^2=16$
• C. $x^2+y^2=25$
• D. $x^2+y^2=\frac{81}{34}$

Answer 1

Answer: (D)

S : $\frac{x^2}{9}-\frac{y^2}{16}=1$
$\Theta C _1$ and $S$ are orthogonally, so tangent of $S$ at $P$ will normal of $C _1$.
So, equation of tangent : $\frac{5 x}{9}-\frac{y}{3}=1 \Rightarrow 5 x-3 y=9$
Tangent will normal to circle so locus of centre of circle will be $5 h -3 k =9$...(1)

Equation of required circle $x ^2+ y ^2= r ^2$
Perpendicular from $(0,0)$ to equation $(1)=$ radius
So, $r =\frac{9}{\sqrt{34}}$
So, equation of circle : $x ^2+ y ^2=\frac{81}{34}$.
Again $C _2$ is locus of centre of circle which touch $S =0$, so normal of $S$ at $P$ passes through centre of circle. So, equation of normal $3 x+5 y=k$ passes through point $P\left(5, \frac{16}{3}\right)$
$\Rightarrow 15+\frac{80}{3}= k $
$\Rightarrow k =\frac{125}{3}, \text { so } C _2=9 x +15 y =125$