Let the lines (2 - i)z = (2 + i)z̅ and (2 + i)z + (i - 2)z̅ - 4i = 0, (here i² = -1) be normal to a circle C. If the line iz + z̅ + 1 + i = 0 is tangent to this circle C, then its radius is :

Solution

Maths sir · 4 steps

Two normals of a circle meet at its centre. The radius is then the distance from it to the tangent.

1. First normal. Put z = x + iy. Then (2-i)z = (2x+y) + i(2y-x) and (2+i)z̅ = (2x+y) + i(x-2y). Equating the imaginary parts gives 2y - x = x - 2y, that is, x = 2y.
2. Second normal. (2+i)z = (2x-y) + i(x+2y) and (i-2)z̅ = (y-2x) + i(x+2y). Adding them and subtracting 4i gives i(2x+4y-4) = 0, so x + 2y = 2.
3. Centre and tangent. Solving x = 2y with x + 2y = 2 gives the centre (1, 1/2). For the tangent, iz + z̅ + 1 + i = (x - y + 1) + i(x - y + 1), so the tangent line is x - y + 1 = 0.

   r = (|1 - 1/2 + 1|)/√(1² + 1²) = (3/2)/√2 = 3/2√2

4. So. The radius is the perpendicular distance from the centre to the tangent, which is 3/2√2. That is option (a).

The idea to keep. Every normal of a circle passes through its centre, so two normals fix the centre. The radius is the perpendicular distance from that centre to any tangent. Convert each complex line to x, y form first.

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