The circle passing through the intersection of the circles,
x² + y² – 6x = 0 and x² + y² – 4y = 0, having its centre on
the line, 2x – 3y + 12 = 0, also passes through the point :

Solution

Maths sir · 3 steps

Write the whole family as S₁ + λ S₂ = 0, then let the line on the centre fix λ.

1. Family of circles. Every circle through the intersection points of S₁: x² + y² - 6x = 0 and S₂: x² + y² - 4y = 0 has the form S₁ + λ S₂ = 0, with λ ≠ -1. Expanding gives (1+λ)(x² + y²) - 6x - 4λ y = 0, so the centre is (3/(1+λ), 2λ/(1+λ)).
2. Centre on the line. Put the centre into 2x - 3y + 12 = 0 and multiply through by 1+λ. This gives 6 - 6λ + 12(1+λ) = 0, so 18 + 6λ = 0 and λ = -3.

   -2(x² + y²) - 6x + 12y = 0 ⇒ x² + y² + 3x - 6y = 0

   Point	x² + y² + 3x - 6y	On the circle?
   (a) (-3, 1)	9 + 1 - 9 - 6 = -5	No
   (b) (1, -3)	1 + 9 + 3 + 18 = 31	No
   (c) (-1, 3)	1 + 9 - 3 - 18 = -11	No
   (d) (-3, 6)	9 + 36 - 9 - 36 = 0	Yes

3. So. The circle is x² + y² + 3x - 6y = 0, with centre (-3/2, 3), which does lie on the line. Of the four points, only (-3, 6) satisfies the equation, so the answer is (d).

The idea to keep. For a circle through the meeting points of two circles, start with S₁ + λ S₂ = 0. The one extra condition, here the centre on a line, fixes λ.

Answer from the NTA final answer key. Solution by PrepWiser.