1.

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

Answer»

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

14f – 3c = -17 … (5)

14f – 3c = -17 

(2) x 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 = \(\frac{-20}{8}\) = \(\frac{-5}{2}\)

Using g = \(\frac{-5}{2}\), f = -1, c = 1 in (1) we get the equation of the circle. 

x2 + y2 + 2(\(\frac{-5}{2}\))x + 2(-1)y + 1 = 0 

x2 + y2 – 5x – 2y + 1 = 0



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