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 isYou need to login to perform this action.
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