Let the centre of a circle, passing through the points (0,0),(1,0) and touching the circle x²+y²=9, be (h, k). Then for all possible values of the coordinates of the centre (h, k), 4(h²+k²) is equal to __________.

Solution

Maths sir · 4 steps

The circle passes through the origin, so its radius is just the distance from O to its centre.

1. Locate the centre. The centre is equidistant from (0,0) and (1,0), so it lies on the perpendicular bisector x = 1/2. Hence h = 1/2, and the radius is r = √(h² + k²), the distance from the centre to the origin.
2. Internal touching. The point (0,0) lies inside x² + y² = 9, so our circle sits inside it and must touch it internally. The distance between centres is then 3 - r. That distance is also √(h² + k²) = r.

   r = 3 - r ⇒ r = 3/2

3. Check the centres. h² + k² = r² = 9/4, so k² = 9/4 - 1/4 = 2 and k = ±√2. Both centres (1/2, ±√2) give the same value.
4. So. 4(h² + k²) = 4 × 9/4 = 9.

The common slip: 2.25. Stopping at h² + k² = 9/4 and forgetting the factor 4 gives 2.25, not 9.

The idea to keep. When a circle passes through the centre of another circle, its radius equals the distance between the centres. Pair that with the touching condition and the radius drops out at once.

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