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 $\left(t,9-t\right)$.

Let the radius of the circle be a.

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

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

Therefore, the equation of the circle is ${\left(x-t\right)}^2+{\left(y-\left(9-t\right)\right)}^2=a^2$. ...(2)

The circle passes through (2, 5).

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

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 ${\left(x-6\right)}^2+{\left(y-3\right)}^2=20$.