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

Answer 1

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