Question

If the line 2x − y + 1 = 0 touches the circle at the point (2, 5) and the centre of the circle lies on the line x + y − 9 = 0. Find the equation of the circle.

Answer 1

According to question, the centre of the required circle lies on the line x + y − 9 = 0.
Let the coordinates of the centre be $$(t,9-t)$$.

Let the radius of the circle be a.
Here, a is the distance of the centre from the line 2x − y + 1 = 0.

$$\therefore a=\left|\frac{2t-9+t+1}{\sqrt{2^2+{(-1)}^2}}\right|=\left|\frac{3t-8}{\sqrt{5}}\right|$$
$$\Rightarrow a^2={\left(\frac{3t-8}{\sqrt{5}}\right)}^2...\left(1\right)$$

Therefore, the equation of the circle is

$${(x-t)}^2+{(y-(9-t))}^2=a^2$$  ...(2)

The circle passes through (2, 5).

∴ $${(2-t)}^2+{(5-(9-t))}^2=a^2$$

$$\Rightarrow {(2-t)}^2+{(5-(9-t))}^2={\left(\frac{3t-8}{\sqrt{5}}\right)}^2\left(Using\left(1\right)\right)$$
$$\Rightarrow 5(2t^2-12t+20)=9t^2+64-48t$$
$$\Rightarrow {(t-6)}^2=0$$
$$\Rightarrow t=6$$

Substituting t = 6 in (1): $$a^2={\left(\frac{10}{\sqrt{5}}\right)}^2$$

Substituting the values of $$a^2$$  and t in equation (2), we find the required equation of circle to be $${(x-6)}^2+{(y-3)}^2=20$$