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
x2 + y2 + (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}) $
$\displaystyle Radius = \sqrt{2}(10 -4 \sqrt{5})$