Exams › JEE Main › Maths

Find the equation of the tangents of slope -1 to the circle |z - (3 + 3i)| = sqrt(2).

1. Im((z - 4 - 4i)/(-1 + i)) = 0
2. Im((z - 2 - 2i)/(-1 + i)) = 0
3. Re((z - 4 - 4i)/(-1 + i)) = 0
4. Re((z - 2 - 2i)/(-1 + i)) = 0

Correct answer: Im((z - 4 - 4i)/(-1 + i)) = 0

Solution

The circle has centre 3+3i, radius sqrt(2). Tangents of slope -1 touch at 4+4i and 2+2i. Writing the line through 4+4i with direction (-1+i) (slope -1) as Im((z-(4+4i))/(-1+i)) = 0 selects the point 4+4i version matching the options.

Related JEE Main Maths questions

• Given z1 = √3 + i√3 and z2 = √3 + i, determine the quadrant in which the complex number z1/z2 lies.
• For the quadratic equation 2(1+i)x² − 4(2−i)x − (5+3i) = 0, the root with the larger absolute value is
• Find the value of ((cosθ+isinθ)⁴)/((cosθ-isinθ)³).
• How many complex numbers z satisfy the equation z³ + (3|z|²)/(z) = 0, given that |z| = √(3)?
• Let z = x + iy be a variable complex number. If arg((z - 1)/(z + 1)) = π/4, then which relation is satisfied?
• For positive values of a, b and c, consider the quadratic equation ax² + bx + c = 0. What can be said about its two roots?

⚔️ Practice JEE Main Maths free + battle 1v1 →