Question

If the line $2x-y+1=0$ is a tangent to the circle $S\equiv 0$ at $P(2,5)$ and the centre of the circle lies on $x-2y=4,$ then the intercept made by the circle $S\equiv 0$ on the $x-$axis is equal to

• A. $2\sqrt{23}$
• B. $2\sqrt{41}$
• C. $2\sqrt{35}$
• D. $2\sqrt{63}$

Answer 1

The correct option is B $2\sqrt{41}$

Equation of normal at $P$ is $x+2y+k=0$
$\Rightarrow 2+2\cdot 5+k=0\Rightarrow k=-12$
$\therefore$ Equation of normal at $P$ is $x+2y-12=0\cdots \left(1\right)$
Centre lies on $x-2y-4=0\cdots \left(2\right)$
Solving $\left(1\right)$ and $\left(2\right),$
we get centre, $C\equiv (8,2)$
Radius, $r=CP=\sqrt{45}$
$\therefore$ Equation of the circle is
$(x-8{)}^2+(y-2{)}^2=45$
$\Rightarrow x^2+y^2-16x-4y+23=0$
Length of $x-$intercept $=2\sqrt{g^2-c}=2\sqrt{41}$