Question

A circle whose centre is in $2^{\text{nd}}$ quadrant touches the line $y=x$ at a point $P$ (in $3^{\text{rd}}$ quadrant) such that $OP=4\sqrt{2}$ unit, where $O$ is the origin. The length of its chord on the line $x+y=0$ is $6\sqrt{2}$ units. If end points of chord is $A,B$ and centre is $C$, then sum of ordinates of $P,A,B,C$ is

Answer 1

Given $OP=4\sqrt{2}$ unit
So possible coordinates of $P$ is $(-4,-4)$
and centre lies in $2^{\text{nd}}$ quadrant
So poosible diagram for the question is
Now $O{P}^2=OA\cdot OB$
$\Rightarrow 32=OA\cdot (AB+OA)\Rightarrow 32=OA(6\sqrt{2}+OA)\Rightarrow OA=2\sqrt{2}\text{units}$
$\Rightarrow OB=8\sqrt{2}$ units
$\therefore B\equiv (-8,8),A\equiv (-2,2)$
Now perpendicular bisector of $AB$ is
$x-y+10=0\cdots \left(1\right)$
and equation of line $\perp OP$ and passing through $P$ is
$x+y+8=0\cdots \left(2\right)$
Intersection of $\left(1\right)$ and $\left(2\right)$ will give the centre of the circle as $(-9,1)$
Hence the required sum $=8+2+1-4=7$