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Question
Find the equation of the circles with centre (2, 3) and passing through the intersection of the lines 3x – 2y – 1 = 0 and 4x + y – 27 = 0
Solution
Centre (2, 3) = (h, k)
Point of intersection
Solve 3x – 2y – 1 = 0 .......(1)
4x + y – 27 = 0 .......(2)
(1) ⇒ 3x – 2y = 1
(2) × 2 ⇒ 8x + 2y = 54
11x = 55
x = 5
Put in (1)
15 – 2y – 1 = 0
14 = 2y
y = 7
Passing-through point is (5, 7)
Equation of circle be (x – h)2 + (y – k)2 = r2 .......(3)
(5 – 2)2 + (7 – 3)2 = r2
32 + 42 = r2
r2 = 25
∴ (3) ⇒ (x – 2)2 + (y – 3)2 = 25
x2 – 4x + 4 + y2 – 6y + 9 – 25 = 0
x2 + y2 – 4x – 6y – 12 = 0
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