JEE Main 2019 (Online) 12th January Morning ShiftQuestion 1
Let $C_1$ and $C_2$ be the centres of the circles $x^{2}+y^{2}-2x-2y-2=0$ and $x^{2}+y^{2}-6x-6y+14=0$ respectively. If $P$ and $Q$ are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral $PC_1QC_2$ is :
Circle
Solution
Formulas used. (1) $x^2+y^2+2gx+2fy+c=0$: centre $(-g,-f)$, radius $\sqrt{g^2+f^2-c}$(2) $d^2=r_1^2+r_2^2\iff$ the circles cut at $90^\circ$ (radii to $P$ are perpendicular).Step 1 - Centres and radii. By (1): $C_1=(1,1)$, $r_1=\sqrt{1+1+2}=2$; $\ C_2=(3,3)$, $r_2=\sqrt{9+9-14}=2$. Step 2 - Angle at $P$. $$C_1C_2^2=(3-1)^2+(3-1)^2=8=r_1^2+r_2^2$$ By (2), $\angle C_1PC_2=90^\circ$, and likewise $\angle C_1QC_2=90^\circ$.Step 3 - Area of the kite. $C_1C_2$ divides $PC_1QC_2$ into two congruent right triangles with legs $2$ and $2$: $$\text{Area}=2\cdot\tfrac12\cdot2\cdot2=4$$ (Check: $P(3,1)$, $Q(1,3)$ satisfy both circles; $\tfrac12\cdot PQ\cdot C_1C_2=\tfrac12\cdot2\sqrt2\cdot2\sqrt2=4$ ✓.) Answer: B