Question

Let $f(x,y)=0$ be the equation of a circle such that $f(0,y)=0$ has equal real roots and $f(x,0)=0$ has two distinct real roots. Let $g(x,y)=0$ be the locus of points ${}^{\prime }{P}^{\prime }$ from where tangents to circle $f(x,y)=0$ make angle $\frac{\pi }{3}$ between them and $g(x,y)=x^2+y^2–5x–4y+c,c\in R$. Let $Q$ be a point from where tangents drawn to circle $g(x,y)=0$ are mutually perpendicular. If $A,B$ are the points of contact of tangents drawn from $Q$ to circle $g(x,y)=0$, then area of triangle $QAB$ is

• A. $\frac{25}{12}$
• B. $\frac{25}{8}$
• C. $\frac{25}{4}$
• D. $\frac{25}{2}$

Answer 1

The correct option is D $\frac{25}{2}$

Given $f(x,y)=0$ is circle.
As $f(0,y)$ has equal roots hence $f(x,y)=0$ touches the $y-$axis and as $f(x,0)=0$ has two distinct real roots hence $f(x,y)=0$ cuts the $x-$axis in two distinct points.


Given $g(x,y)=x^2+y^2–5x–4y+c$
$\text{centre}=(\frac{5}{2},2),\text{radius}=\sqrt{\frac{25}{4}+4-c}$

From figure, radius of $g(x,y)=$ twice the radius of $f(x,y)=0$ and $r=\frac{5}{2}$
$\therefore$ Radius of $g(x,y)=5$
Hence $\frac{25}{4}+4-c=25\Rightarrow c=-\frac{59}{4}$

$\therefore$ Equation of $g(x,y)$ is
$x^2+y^2-5x-4y-\frac{59}{4}=0$

Area of $\Delta QAB=\frac{1}{2}\times 5\times 5=\frac{25}{2}$