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

Deepak Kumar Shringi
6 years ago
Samyak Jain
333 Points
5 years ago
Let equation of circle be x2ย + y2ย + 2gx + 2fy + c = 0,
where centre is (โ g,โ f) and radius isย $\dpi{100} \sqrt{g^{2} + f^{2} - c}$
x โ 2a = 0 is a tangent to the circle.
$\dpi{80} \therefore$ย perpendicular distance from centre to the line is equal to the radius of the circle.
|โ g โ 2a| =ย $\dpi{100} \sqrt{g^{2} + f^{2} - c}$ย  ย  ย  ย  ย  ย  ย  โฆ.(1)
The circle cuts given circle orthogonally.
$\dpi{80} \therefore$ย 2g1g2 + 2f1f2 = c1 + c2
$\dpi{80} \Rightarrow$ย 2(-g)(0) + 2(-f)(0) = c โ a2ย ย $\dpi{80} \Rightarrow$ย  c =ย a2
(1) becomes |g + 2a| =ย $\dpi{80} \sqrt{g^{2} + f^{2} - a^{2}}$
Square both sides.
g2ย + 4ag + 4a2ย = g2ย + f2ย โ a2
f2ย = 4ag + 5a2
Replace โg by x and โf by yย  i.e.ย  g by โx and f by โy
$\dpi{80} \therefore$ย y2ย = โ 4ax + 5a2ย  ย i.e.
y2ย + 4ax โ 5a2ย = 0ย ย is the required locus of the centres of the circles.