question_answer

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 \[P{{C}_{1}}Q{{C}_{2}}\] is [JEE Main Online Paper Held On 12-Jan-2019 Morning]

A) 8                                             

B) 4    

C) 6                                 

D)   9

Correct Answer: B

Solution :

Let\[{{S}_{1}}\equiv {{x}^{2}}+{{y}^{2}}-2x-2y-2=0\] \[{{S}_{2}}\equiv {{x}^{2}}+{{y}^{2}}-6x-6y+14=0\] \[\therefore \]\[{{C}_{1}}\equiv (1,1)\]and\[{{C}_{2}}\equiv (3,3).\] Radius of \[{{S}_{1}},P{{C}_{1}}=\sqrt{1+1+2}=2\] Radius of\[{{S}_{2}},P{{C}_{2}}=\sqrt{9+9-14}=2\] \[\therefore \]\[P{{C}_{1}}=Q{{C}_{1}}=P{{C}_{2}}=Q{{C}_{2}}=2\] Now, \[2{{g}_{1}}{{g}_{2}}+2{{f}_{1}}{{f}_{2}}\] \[=2\times 3+2\times 3=6+6=12\] and \[{{c}_{1}}+{{c}_{2}}=14-2=12\] Here, \[2{{g}_{1}}{{g}_{2}}+2{{f}_{1}}{{f}_{2}}={{c}_{1}}+{{c}_{2}}\] \[\therefore \]Both circles are orthogonal. So, \[P{{C}_{1}}Q{{C}_{2}}\]is a square. \[\therefore \]Area of \[P{{C}_{1}}Q{{C}_{2}}=2\times 2=4sq.\]units