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 res…

Question

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:

Accepted Answer

Centres $C_1=(1,1)$, $C_2=(3,3)$; the circles are orthogonal since $2(1)(3)+2(1)(3)=12=(-2)+14$ (matching $2g_1g_2+2f_1f_2=c_1+c_2$).

For orthogonal circles, $PC_1QC_2$ is a kite with $C_1P\perp PC_2$, and area $=2\cdot	ext{area}(	riangle PC_1C_2)=2\cdot\dfrac12\cdot2\cdot2=4.$

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