JEE PYQ: Circle Question 42
Question 42 - 2019 (12 Jan Shift 2)
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:
(1) $8$
(2) $6$
(3) $9$
(4) $4$
Show Answer
Answer: (4)
Solution
$C_1(1,1), r_1 = 2$; $C_2(3,3), r_2 = 2$. Circles intersect orthogonally: $2g_1g_2 + 2f_1f_2 = c_1 + c_2 \Rightarrow 2(-1)(-3) + 2(-1)(-3) = -2 + 14 = 12$. Yes, orthogonal. Area $= 2 \times \frac{1}{2} r_1 r_2 = 4$.
BETA
