Question

If the lines 2x – 3y – 5 = 0 and 3x – 4y = 7 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

Answer: (C)

x² + y² – 2x + 2y – 47 = 0

Answer 2

The centre of the required circle lies at the point of intersection of the lines 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, πr² = 154

⇒ $\frac{22}{7}$r² = 154 ⇒ r = 7.

Hence, the equation of the required circle is

(x-1)² + (y+1)² = 7² ⇒ x²+y²-2x+2y-47 = 0