55. What is the equation of the circle which
passes through the points (3,-2) and
(-2, 0) and having its centre on the line
2x - Y-3 = 0?
(a) x2 + y2 + 3x + 2 = 0
(b) x2 + y2 + 3x +12y + 2 = 0
(c) x2 + y2 + 2x = O
(d) x2 + y2 = 5
Answers
Step-by-step explanation:
Given: The line 2x – y = 3 … (1) The points (3, -2), (-2, 0) By using the standard form of the equation of the circle: x2 + y2 + 2ax + 2by + c = 0 ….. (2) Let us substitute the centre (-a, -b) in equation (1) we get, 2(- a) – (- b) = 3 -2a + b = 3 2a – b + 3 = 0…… (3) Now Substitute the given points (3, -2) in equation (2), we get 32 + (- 2)2 + 2a(3) + 2b(- 2) + c = 0 9 + 4 + 6a – 4b + c = 0 6a – 4b + c + 13 = 0….. (4) Substitute the points (-2, 0) in equation (2), we get (- 2)2 + 02 + 2a(- 2) + 2b(0) + c = 0 4 + 0 – 4a + c = 0 4a – c – 4 = 0….. (5) By simplifying the equations (3), (4) and (5) we get, a = 3/2, b = 6, c = 2 Again by substituting the values of a, b, c in (2), we get x2 + y2 + 2 (3/2)x + 2 (6)y + 2 = 0 x2 + y2 + 3x + 12y + 2 = 0 ∴ The equation of the circle is (b) x2 + y2 + 3x + 12y + 2 = 0.