# 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

Deepak Kumar Shringi
4 years ago
Samyak Jain
333 Points
4 years ago
Let equation of circle be x2 + y2 + 2gx + 2fy + c = 0,
where centre is (– g,– f) and radius is $\dpi{100} \sqrt{g^{2} + f^{2} - c}$
x – 2a = 0 is a tangent to the circle.
$\dpi{80} \therefore$ perpendicular distance from centre to the line is equal to the radius of the circle.
|– g – 2a| = $\dpi{100} \sqrt{g^{2} + f^{2} - c}$              ….(1)
The circle cuts given circle orthogonally.
$\dpi{80} \therefore$ 2g1g2 + 2f1f2 = c1 + c2
$\dpi{80} \Rightarrow$ 2(-g)(0) + 2(-f)(0) = c – a2  $\dpi{80} \Rightarrow$  c = a2
(1) becomes |g + 2a| = $\dpi{80} \sqrt{g^{2} + f^{2} - a^{2}}$
Square both sides.
g2 + 4ag + 4a2 = g2 + f2 – a2
f= 4ag + 5a2
Replace –g by x and –f by y  i.e.  g by –x and f by –y
$\dpi{80} \therefore$ y2 = – 4ax + 5a2   i.e.
y2 + 4ax – 5a2 = 0  is the required locus of the centres of the circles.