Question

Find the equation of the circle which has its centre at the point (3, 4) and touches the straight line 5x + 12y − 1 = 0.

Answer 1

It is given that the centre is at the point (3, 4).
Let the equation of the circle be

\[\left( x - h \right)^2 + \left( y - k \right)^2 = a^2\]

∴ Equation of the required circle =

\[\left( x - 3 \right)^2 + \left( y - 4 \right)^2 = a^2\]...(1)

Also, the circle touches the straight line 5x + 12y − 1 = 0.

\[\therefore a = \left| \frac{5\left( 3 \right) + 12\left( 4 \right) - 1}{\sqrt{5^2 + {12}^2}} \right| = \left| \frac{62}{13} \right|\]
\[ \Rightarrow a^2 = \left| \frac{5\left( 3 \right) + 12\left( 4 \right) - 1}{13} \right| = \frac{3844}{169}\]

So, from equation (1), we have:

\[\left( x - 3 \right)^2 + \left( y - 4 \right)^2 = \frac{3844}{169}\]

\[\Rightarrow x^2 + y^2 - 6x - 8y = \frac{3844}{169} - 25\]

\[\Rightarrow 169\left( x^2 + y^2 - 6x - 8y \right) + 381 = 0\]

Hence, the required equation of the circle is

\[169\left( x^2 + y^2 - 6x - 8y \right) + 381 = 0\]