Question

From a point $P$ on the normal $y=x+c$ of the circle $x^2+y^2-2x-4y+5-k^2=0,$ two tangents are drawn to the same circle touching it at point $B$ and $C.$ If the area of the quadrilateral $OBPC$ (where $O$ is the centre of the cicrle ) is $36$ sq. units and ponit $P$ is at a distance of $|k|(\sqrt{2}-1)$ from the circle, then the possible value(s) of $k$ is/are

• A. $6$
• B. $6\sqrt{2}$
• C. $-6$
• D. $4\sqrt{2}$

Answer 1

Answer: (A), (C)

The correct options are
A $6$
C $-6$


Since, the line $y=x+c$ is normal to the given circle.
$\Rightarrow c=1$

Equation of the line is $y=x+1...\left(1\right)$
Radius of the circle $=|k|$
Given $AP=|k|(\sqrt{2}-1)$
$\Rightarrow OP=\sqrt{2}|k|$
$\Rightarrow PC=|k|\because OC=|k|$

Now, area of the quadrilateral $OBPC=36$
$\Rightarrow 2\times \frac{1}{2}|k{|}^2=36$
$(\because$ area of $OCP=OBP=\frac{1}{2}|k{|}^2)$
$\Rightarrow k=\pm 6$