Circle And System Of Circles Question 41
Question: The locus of centre of a circle passing through (a, b) and cuts orthogonally to circle $ x^{2}+y^{2}=p^{2} $ , is
[IIT 1988; AIEEE 2005]
Options:
A) $ 2ax+2by-(a^{2}+b^{2}+p^{2})=0 $
B) $ 2ax+2by-(a^{2}-b^{2}+p^{2})=0 $
C) $ x^{2}+y^{2}-3ax-4by+(a^{2}+b^{2}-p^{2})=0 $
D) $ x^{2}+y^{2}-2ax-3by+(a^{2}-b^{2}-p^{2})=0 $
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Answer:
Correct Answer: A
Solution:
Let equation of circle be $ x^{2}+y^{2}+2gx+2fy+c=0 $ with $ x^{2}+y^{2}=p^{2} $ cutting orthogonally, we get $ 0+0=+c-p^{2} $ or $ c=p^{2} $ and passes through (a, b),
we get $ a^{2}+b^{2}+2ga+2fb+p^{2}=0 $ or $ 2ax+2by-(a^{2}+b^{2}+p^{2})=0 $
Required locus as centre $ (-g,\ -f) $ is changed to (x, y).
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