1.

If the lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 sq. units, then find the equation of the circle. (A) x2 + y2 – 2x + 2y = 40(B) x2 + y2 – 2x – 2y = 47 (C) x2 + y2 – 2x + 2y = 47 (D) x2 + y2 – 2x – 2y = 40

Answer»

Correct option is (C) x2 + y2 – 2x + 2y = 47 

Centre of circle = Point of intersection of diameters.

Solving 2x – 3y = 5 and 3x – 4y = 7, we get 

x = 1, y = -1 

Centre of the circle C(h, k) = C(1, -1) 

∴ Area = 154 

πr2 = 154

22/7 x r2 = 154

r2 = 154 x 22/7 = 49

∴ r = 7 

equation of the circle is 

(x – 1)2 + (y + 1)2 = 72 

x2 + y2 – 2x + 2y = 47



Discussion

No Comment Found

Related InterviewSolutions