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.

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

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).

∴ \[\left( 2 - t \right)^2 + \left( 5 - \left( 9 - t \right) \right)^2 = a^2\]

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

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 \[\left( x - 6 \right)^2 + \left( y - 3 \right)^2 = 20\]