Question

Find the equation of the circle passing through the points (0, 1), (4, 3) and (1, -1).

Answer 1

Let the required of the circle be x² + y² + 2gx + 2fy + c = 0 ……… (1)

It passes through (0, 1)

0 + 1 + 2g(0) + 2f(1) + c = 0

1 + 2f + c = 0

2f + c = -1 …….. (2)

Again the circle (1) passes through (4, 3)

4² + 3² + 2g(4) + 2f(3) + c = 0

16 + 9 + 8g + 6f + c = 0

8g + 6f + c = -25 …….. (3)

Again the circle (1) passes through (1, -1)

1² + (-1)² + 2g(1) + 2f(-1) + c = 0

1 + 1 + 2g – 2f + c = 0

2g – 2f + c = -2 ……… (4)

8g + 6f + c = -25

(4) × 4 subtracting we get, 8g – 8f + 4c = -8

14f – 3c = -17 ………. (5)

14f – 3c = -17

(2) × 3 ⇒ 6f + 3c = -3

Adding we get 20f = -20

f = -1

Using f = -1 in (2) we get, 2(-1) + c = -1

c = -1 + 2

c = 1

Using f = -1, c = 1 in (3) we get

8g + 6(-1)+1 = -25

8g – 6 + 1 = -25

8g – 5 = -25

8g = -20

g = `(-20)/8 = (-5)/2`

using g = `(-5)/2`, f = -1, c = 1 in (1) we get the equation of the circle.

x² + y² + 2`((-5)/2)x + 2(-1)y + 1 = 0`

x² + y² – 5x – 2y + 1 = 0