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Question
Find the equation of the circle passing through the points (5, 7), (6, 6) and (2, −2)
Solution
Let P (h, k) be the centre of the circle and A (5, 7), B (6, 6) and C (2, – 2) be the points on the circle.
Then PA = PB = PC
PA= PB gives: `sqrt(("h" - 5)^2 + ("k" - 7)^2`
= `sqrt(("h" - 6)^2 + ("k" - 6)^2`
On squaring both sides, we get,
h2 – 10h + 25 + k2 – 14k + 49
= h2 – 12h + 36 + k2 – 12k + 36
∴ 2h – 2k = – 2
∴ h – k = – 1 ... (1)
PA = PC gives: `sqrt(("h" - 5)^2 + ("k" - 7)^2`
= `sqrt(("h" - 2)^2 + ("k" + 2)^2`
On squaring both sides, we get,
h2 – 10h + 25 + k2 – 14k + 49 = h2 – 4h + 4 + k2 + 4k + 4
∴ – 6h – 18k = – 66
∴ h + 3k = 11 ...(2)
Subtracting equation (1) from (2), we get,
4k = 12
∴ k = 3
∴ from (1), h – 3 = – 1
∴ h = 2
∴ the centre P is (2, 3)
∴ radius = PA
= `sqrt((5 - 2)^2 + (7 - 3)^2`
= `sqrt(9 + 16)`
= 5
∴ the equation of the required circle is
(x – 2)2 + (y – 3)2 = 52
∴ x2 – 4x + 4 + y2 – 6y + 9 = 25
∴ x2 + y2 – 4x – 6y – 12 = 0.
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