Find the equation of the circle which passes through (2, -2) and (3, 4

1.	Find the equation of the circle which passes through (2, -2) and (3, 4) and whose centre lies on the line x + y = 2.
Answer» Let, the equation of the circle with centre (h, k) and radius ‘r’ be (x − h)² + (y − k)² = r² … (1) Since the circle passes through (2, −2) and (3, 4). (2 − h)² + (−2 − k)² = r² … (2) And (3 − h)² + (4 − k)² = r² … (3) (2) and (3) ⇒ (2 − h)²+ (2 + k)² = (3 − h)² + (4 − k)² ⇒ 4 − 4h + h² + 4 + 4k + k² = 9 − 6h + h² + 16 − 8k + k² ⇒ −4h + 4k+ 8 + 6h + 8k − 25 = 0 ⇒ 2h + 12k − 17 = 0 … (4) Also, centre (h, k) lies on x + y = 2 ∴h + k = 2 … (5) (4) − 2 × (5) ⇒ 10k − 17 = −4 ⇒ \(k=\frac{13}{10}\) (5) ⇒ k = 1.3 ⇒ h= 0.7 ∴ (1) ⇒ (x − 0.7)² + (y − 1.3)² = r² … (6) (2) ⇒ (2 − 0.7)² + (−2 − 1.3)² = r² ⇒ r² = 12.58 ∴ (6) ⇒ (x − 0.7)²+ (y − 13)² = 12.58, is the required equation of the circle.

Find the equation of the circle which passes through (2, -2) and (3, 4) and whose centre lies on the line x + y = 2.

Answer»

Let, the equation of the circle with centre (h, k) and radius ‘r’ be

(x − h)² + (y − k)² = r² … (1)

Since the circle passes through (2, −2) and (3, 4).

(2 − h)² + (−2 − k)² = r² … (2)

And (3 − h)² + (4 − k)² = r² … (3)

(2) and (3) ⇒ (2 − h)²+ (2 + k)² = (3 − h)² + (4 − k)²

⇒ 4 − 4h + h² + 4 + 4k + k² = 9 − 6h + h² + 16 − 8k + k²

⇒ −4h + 4k+ 8 + 6h + 8k − 25 = 0

⇒ 2h + 12k − 17 = 0 … (4)

Also, centre (h, k) lies on x + y = 2

∴h + k = 2 … (5)

(4) − 2 × (5) ⇒ 10k − 17 = −4

⇒ \(k=\frac{13}{10}\)

(5) ⇒ k = 1.3 ⇒ h= 0.7

∴ (1) ⇒ (x − 0.7)² + (y − 1.3)² = r² … (6)

(2) ⇒ (2 − 0.7)² + (−2 − 1.3)² = r²

⇒ r² = 12.58 ∴ (6)

⇒ (x − 0.7)²+ (y − 13)² = 12.58, is the required equation of the circle.