If a line, y = mx + c is a tangent to the circle, (x – 3)² + y² = 1 and it is perpendicular to a line L₁, where L₁ is the tangent to the circle, x² + y² = 1 at the point ( 1/√2,1/√2 ), then :

Solution

Maths sir · 4 steps

Get the slope from the first circle, then make the line sit at distance 1 from the second centre.

1. Find L1. The tangent to x² + y² = 1 at (x₁, y₁) is x x₁ + y y₁ = 1. At (1/√2, 1/√2) this gives x + y = √2, so L₁ has slope -1.
2. Fix the slope. The line y = mx + c is perpendicular to L₁, so m × (-1) = -1, which gives m = 1. The line is x - y + c = 0.
3. Tangency condition. The circle (x-3)² + y² = 1 has centre (3, 0) and radius 1. The line touches it when its perpendicular distance from the centre equals the radius.

   (|3 - 0 + c|)/√(1² + 1²) = 1 ⇒ (c + 3)² = 2

4. So. Expanding, c² + 6c + 9 = 2, that is c² + 6c + 7 = 0. This is option (a).

The idea to keep. For a line to touch a circle, set the distance from the centre to the line equal to the radius. Perpendicular lines have slopes whose product is -1.

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