The locus of the complex number z satisfying | z - 3i z + 3i | = 2 is a circle C . A pair of tangents are drawn from the origin O to C , touching it at points P and Q . If K is the center of the circle C , then the area of the quadrilateral OPKQ (in square units) is:

Step-by-step solution

Let z = x + iy z = x - iy . The given equation is |x - iy - 3i| = 2|x - iy + 3i| . |x - i(y + 3)| = 2|x - i(y - 3)| Squaring both sides, we get: x^2 + (y + 3)^2 = 4[x^2 + (y - 3)^2] x^2 + y^2 + 6y + 9 = 4x^2 + 4y^2 - 24y + 36 3x^2 + 3y^2 - 30y + 27 = 0 x^2 + y^2 - 10y + 9 = 0 This represents a circle with center K(0, 5) and radius r = 0^2 + (-5)^2 - 9 = 25 - 9 = 4 . The distance from the origin O(0, 0) to the center K(0, 5) is d = 5 . The length of the tangent OP from the origin to the circle is given by OP = d^2 -