the locus of centres of all circles which touch the line x=2a and cut the circle x^2+y^2=a^2 orthogonally is

Question

Deepak Kumar Shringi · Answer

Samyak Jain · Answer

Let equation of circle be x 2 + y 2 + 2gx + 2fy + c = 0, where centre is (– g,– f) and radius is x – 2a = 0 is a tangent to the circle. perpendicular distance from centre to the line is equal to the radius of the circle. |– g – 2a| = ….(1) The circle cuts given circle orthogonally. 2g1g2 + 2f1f2 = c1 + c2 2(-g)(0) + 2(-f)(0) = c – a 2 c = a 2 (1) becomes |g + 2a| = Square both sides. g 2 + 4ag + 4a 2 = g 2 + f 2 – a 2 f 2 = 4ag + 5a 2 Replace –g by x and –f by y i.e. g by –x and f by –y y 2 = – 4ax + 5a 2 i.e. y 2 + 4ax – 5a 2 = 0 is the required locus of the centres of the circles.

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the locus of centres of all circles which touch the line x=2a and cut the circle x^2+y^2=a^2 orthogonally is

the locus of centres of all circles which touch the line x=2a and cut the circle x^2+y^2=a^2 orthogonally is

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