Question

If the lines 2x − 3y = 5 and 3x − 4y = 7 are the diameters of a circle of area 154 square units, then obtain the equation of the circle.

Answer 1

We have Area of circle = 154

\[\pi r^2 = 154\]
\[ \Rightarrow r^2 = 49\]

The intersection of two lines will give us the centre of the circle.
Solving 2x − 3y = 5 and 3x − 4y = 7 we get
x = 1 and y = −1
Now, the equation of the circle is given by

\[\left( x - h \right)^2 + \left( y - k \right)^2 = r^2 \]
\[ \Rightarrow \left( x - 1 \right)^2 + \left( y + 1 \right)^2 = 49\]
\[ \Rightarrow x^2 + y^2 - 2x + 2y - 47 = 0\]