Question

Find the equation of the circle passing through (1,1),(2,-1),(3,2)

Answer 1

Step-by-step explanation:

Let, the equation of the circle be

x² + y² + ax + by + c = 0 ...(i)

Since, (i) no. circle passes through the given points, they satisfy the equation (i).

Then,

1² + 1² + a + b + c = 0

➨ a + b + c = - 2 ...(ii)

2² + (-1)² + 2a - b + c = 0

➨ 2a - b + c = - 5 ...(iii)

3² + 2² + 3a + 2b + c = 0

➨ 3a + 2b + c = - 13 ...(iv)

Now, adding (ii) and (iii), we get

3a + 2c = - 7 ...(v)

Again, {(ii) × 2} - (iv) ➨

2a + 2b + 2c - 3a - 2b - c = - 4 + 13

➨ - a + c = 9

➨ - 3a + 3c = 27 ...(vi)

Adding (v) and (vi), we get

5c = 20

➨ c = 4

Putting c = 4 in (v), we get

3a + (2 × 4) = - 7

➨ 3a = - 7 - 8

➨ 3a = - 15

➨ a = - 5

Now, putting a = - 5 and c = 4 in (ii), we get

- 5 + b + 4 = - 2

➨ b = - 2 + 1

➨ b = - 1

Putting a = - 5, b = - 1 and c = 4 in (i), we get the required circle as

x² + y² - 5x - y + 4 = 0

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