Let $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ be the centres of the circles $\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}-2 \mathrm{y}-2=0$ and $\mathrm{x}^{2}+\mathrm{y}^{2}-6 \mathrm{x}- 6 y+14=0$ respectively. If P and Q are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral $\mathrm{PC}_{1} \mathrm{QC}_{2}$ is:

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Circle

Let

C_{1}

and

C_{2}

be the centres of the circles

x^{2} + y^{2} - 2 x - 2 y - 2 = 0

and

x^{2} + y^{2} - 6 x - 6 y + 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_{1} QC_{2}

is:

Answer: (D)

The first circle has centre

(1, 1)

and radius

2

; the second has centre

(3, 3)

and radius

2

. The centres are

(3 - 1)^{2} + (3 - 1)^{2} = 22

apart, and since

(22)^{2} = 8 = 2^{2} + 2^{2}

, the two radii to an intersection point meet at a right angle, so the circles are orthogonal. The figure

P C_{1} Q C_{2}

then has all four sides equal to the radius

2

, making it a rhombus; one diagonal is

C_{1} C_{2} = 22

, and from the side length the other diagonal

P Q

also works out to

22

. Its area is

\frac{1}{2} (22) (22) = 4

square units.

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option