question_answer

If a circle passes through the point (a, b) and  cuts  the  circle\[{{x}^{2}}+{{y}^{2}}={{p}^{2}}\]orthogonally, then the equation of the locus of its centre is     AIEEE  Solved  Paper-2005

A) \[2ax+2by-({{a}^{2}}+{{b}^{2}}+{{p}^{2}})=0\]

B) \[{{x}^{2}}+{{y}^{2}}-2ax-3by+({{a}^{2}}-{{b}^{2}}-{{p}^{2}})=0\]

C) \[2ax+2by-({{a}^{2}}-{{b}^{2}}+{{p}^{2}})=0\]

D) \[{{x}^{2}}+{{y}^{2}}-3ax-4by+({{a}^{2}}+{{b}^{2}}-{{p}^{2}})=0\]

Correct Answer: A

Solution :

Two circlesand cuts orthogonally, then Let the circle be It cuts circleorthogonally.           So,and it passes through (a, b). So, locus of is \[\Rightarrow \,\,\,2ax\,+2by\,-({{a}^{2}}+{{b}^{2}}+{{p}^{2}})=0\] Alternate Solution Let the centre of required circle be. This circle cut the circleorthogonally. The centre and radius of circle are (0, 0)and P, respectively.    Hence, the locus of centre is