Let the lines y + 2x = √11 + 7√7 and 2y + x = 2√11 + 6√7 be normal to a circle C:(x - h)² + (y - k)² = r². If the line √11 y - 3x = 5√77/3 + 11 is tangent to the circle C, then the value of (5h - 8k)² + 5r² is equal to __________.

Solution

Maths sir · 4 steps

Every normal passes through the centre, so where the two normals meet is exactly (h, k).

1. Centre from normals. Double the first line to get 2y + 4x = 2√11 + 14√7. Subtract the second line from it: 3x = 8√7, so h = 8√7/3. Then k = √11 + 7√7 - 16√7/3 = √11 + 5√7/3.
2. First term. 5h - 8k = 40√7/3 - 8√11 - 40√7/3 = -8√11. So (5h - 8k)² = 64 × 11 = 704.
3. Radius from tangent. The radius is the perpendicular distance from the centre to the tangent √11 y - 3x - 5√77/3 - 11 = 0. Here √11 k = 11 + 5√77/3 and 3h = 8√7, so the numerator reduces to -8√7.

   r = (|-8√7|)/√(11 + 9) = 8√7/√20 ⇒ r² = 448/20 = 112/5

4. So. 5r² = 112, and (5h - 8k)² + 5r² = 704 + 112 = 816.

The idea to keep. Two normals meet at the centre, and the distance from the centre to a tangent is the radius. Keep the surds unexpanded, because the messy terms usually cancel.

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