If one of the diameters of the circle x² + y² - 2√2 x - 6√2 y + 14 = 0 is a chord of the circle (x - 2√2 )² + (y - 2√2 )² = r², then the value of r² is equal to ____________.

Solution

Maths sir · 5 steps

A diameter of the first circle as a chord means the common chord must pass through its centre.

1. First circle. Its centre is C₁ = (√2, 3√2). Its radius squared is r₁² = 2 + 18 - 14 = 6.
2. Second circle. Expand (x-2√2)² + (y-2√2)² = r² to get x² + y² - 4√2 x - 4√2 y + 16 - r² = 0, with centre C₂ = (2√2, 2√2).
3. Common chord. The ends of the diameter lie on both circles, so that diameter is the common chord. It lies along the radical axis S₁ - S₂ = 0, and this line must pass through C₁.

   S₁ - S₂ = 2√2 x - 2√2 y - 2 + r² = 0

4. Put in C1. At (√2, 3√2) we get 4 - 12 - 2 + r² = 0. As a check, C₁ is the midpoint of the chord, so r² = C₁C₂² + r₁² = 4 + 6.
5. So. r² = 10.

The common slip: 28. Using the full diameter (2√6)² = 24 in place of the half-chord r₁² = 6 gives 4 + 24 = 28, not 10.

The idea to keep. When a diameter of one circle is a chord of another, the radical axis passes through the first centre. Equivalently, r² = d² + r₁², where d is the distance between the centres.

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