Find the equation of the circle which passes through (3, – 2), (- 2,

1.	Find the equation of the circle which passes through (3, – 2), (- 2, 0) and has its centre on the line 2x – y = 3.
Answer» 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 x² + y² + 3x + 12y + 2 = 0.

Find the equation of the circle which passes through (3, – 2), (- 2, 0) and has its centre on the line 2x – y = 3.

Answer»

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 x² + y² + 3x + 12y + 2 = 0.