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 : $\frac{5x}{9}-\frac{y}{3}=1\Rightarrow 5x-3y=9$
Tangent will normal to circle so locus of centre of circle will be $5h-3k=9$...(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 $3x+5y=k$ passes through point $P\left(5,\frac{16}{3}\right)$
$\Rightarrow 15+\frac{80}{3}=k$
$\Rightarrow k=\frac{125}{3},\text{ so }{C}_2=9x+15y=125$