Saved Bookmarks
| 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 |
|