Q: Find the radius of the smallest circle which touches the line 3x – y = 6 at (1 , -3) and also touches the line y = x. Compute the radius approximately.

Sol. The circle is (x – 1)^{2} + (y + 3)2 + l(3x – y – 6) = 0 …(i)

which touches line 3x – y = 6 at (1, -3)

It can be written as

x^{2} + y^{2} + (3λ – 2)x + (6 – λ)y + (10 – 6λ) = 0 …(ii)

This touches y = x

Radius = length of perpendicular from centre $(1-\frac{3\lambda}{2} , \frac{\lambda}{2}-3) $ on x – y = 0

$\displaystyle (\frac{3\lambda -2}{2})^2 + (\frac{6-\lambda}{2})^2 -(10 -\lambda) = (\frac{1-\frac{3}{2}\lambda -\frac{\lambda}{2}+3}{\sqrt{2}})^2$

$\displaystyle \lambda = – 8 + 4 \sqrt{5} $

$\displaystyle (Radius)^2 = \frac{10 \lambda^2}{4} = \frac{5 \lambda^2}{2}$

$\displaystyle Radius = \frac{\sqrt{5}}{\sqrt{2}} (- 8 + 4 \sqrt{5}) $

$\displaystyle Radius = \sqrt{2}(10 -4 \sqrt{5})$