Question

The lines 2x – 3y – 5 = 0 and 3x – 4y – 7 = 0 are diameters of a circle of area 154 square units then the equation of the circle is

(a) x² + y² + 2x – 2y – 62 = 0

(b) x² + y² + 2x – 2y – 47 = 0 

(c) x² + y² – 2x + 2y – 47 = 0

(d) x² + y² – 2x + 2y – 62 = 0

Answer 1

The correct option (c) x² + y² – 2x + 2y – 47 = 0

Explanation:

Area of circle = πr² = 154 ⇒ r² = [(154 × 7)/22] = 49

∴          r = 7

also     2x – 3y – 5 = 0 & 3x – 4y – 7 = 0 are diameter.

∴ Point of intersection is the centre i.e. (1, – 1)

∴ Equation of circle: (x – 1)² + (y + 1)² = 72

∴   x² + y² – 2x + 2y – 47 = 0.