The equation of a circle is Re(z²) + 2(Im(z))² + 2Re(z) = 0, where z = x + iy. A line which passes through the center of the given circle and the vertex of the parabola, x² - 6x - y + 13 = 0, has y-intercept equal to ______________.

Solution

Maths sir · 3 steps

Turn both curves into x and y first; then it is just a line through two points.

1. Find the circle. With z = x + iy, z² = x² - y² + 2ixy, so Re(z²) = x² - y². Adding 2y² and 2x gives x² + y² + 2x = 0, that is (x+1)² + y² = 1, so the centre is (-1, 0).
2. Find the vertex. Write the parabola as y = x² - 6x + 13 = (x - 3)² + 4. Its vertex is at (3, 4).

   m = (4 - 0)/(3 - (-1)) = 1, y - 0 = 1 (x + 1) ⇒ y = x + 1

3. So. Put x = 0 in y = x + 1. The y-intercept is 1.

The common slip: -2. Reading the centre of x² + y² + 2x = 0 as (1, 0) instead of (-1, 0) gives the line y = 2x - 2 and an intercept of -2.

The idea to keep. For locus questions in complex numbers, substitute z = x + iy at once and complete the square. The centre and radius then come straight out of the Cartesian equation.

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