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Q. The lines $2x - 3y - 5 = 0$ and $3x - 4y = 7$ are diameters of a circle of area 154 sq units, then the equation of the circle is

VITEEEVITEEE 2012

Solution:

The centre of the required circle lies at the intersection of $2x - 3y - 5 = 0$ and $3x - 4y - 7 = 0$.
Thus, the coordinates of the centre are $(1, -1)$.
Let r be the radius of the circle.
Then $\pi r^2 = 154$
$\Rightarrow \, \frac{22}{7} r^2 = 154 $
$ \Rightarrow r = 7 $
Hence, the equation of required circle is
$(x -1)^2 + (y + 1)^2 = 7^2$
$\Rightarrow \, x^2 + y^2 - 2x + 2y - 47 = 0$