Let C be a circle passing through the points A(2, -1) and B(3, 4). The line segment AB s not a diameter of C. If r is the radius of C and its centre lies on the circle (x - 5)² + (y - 1)² = 13/2, then r² is equal to :

Solution

Maths sir · 4 steps

The centre is equidistant from A and B, so put it on the perpendicular bisector of AB first.

1. Perpendicular bisector. The midpoint of AB is (5/2, 3/2) and AB has direction (1, 5). The bisector is (x - 5/2) + 5(y - 3/2) = 0, that is x + 5y = 10.
2. Meet the given circle. Put x = 10 - 5y into (x-5)² + (y-1)² = 13/2. Since x - 5 = 5(1 - y), the left side becomes 26(y-1)².

   26(y-1)² = 13/2 ⇒ (y-1)² = 1/4 ⇒ y = 3/2 or 1/2

3. Reject one centre. y = 3/2 gives the centre (5/2, 3/2), the midpoint of AB. That would make AB a diameter, which is not allowed. So the centre is (15/2, 1/2).
4. So. r² = (15/2 - 2)² + (1/2 + 1)² = 121/4 + 9/4 = 65/2. B gives the same value: 81/4 + 49/4 = 65/2. The answer is (b).

The idea to keep. Any circle through two fixed points has its centre on their perpendicular bisector. Intersect that line with the extra condition, then use the given restriction to pick the right centre.

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