Question

If a circle passes through the point (a,b) and cuts the circle x²+y² = K² orthogonally then the equation of the locus of its centre is

• A. 2ax+2by-(a²+b²+K²) = 0
• B. 2ax+2by-(a²-b²+K²) = 0
• C. x²+y²-3ax-4by+(a²+b²-K²) = 0
• D. x²+y²-2ax-3by+(a²-b²-K²) = 0

Answer 1

Answer: (A)

2ax+2by-(a²+b²+K²) = 0

Answer 2

Let the equation of the circle through (a, b) be

x²+y²+2gx+2fy+c = 0 …(1)

⇒ a² + b² + 2 ag + 2 bf + c = 0.…(2)

Since (1) cuts x² + y² = k² orthogonally,

2g. 0 + 2f. 0 = c – k² ⇒ c = k².

Putting c = k² in (2), we get 2 ag + 2 bf + (a²+b²+k²) = 0

so that the locus of the centre (-g, -f) is

– 2ax – 2by + (a² + b² + k²) = 0 or

2 ax+2 by – (a² + b² + k² ) = 0