The locus of the mid points of the chords of the circle $x^2 + y^2 - ax - by = 0$ which subtend a right angle at $(a/2, b/2)$ is:
Step-by-Step Solution
Key Concept: The locus of the center of circles passing through two fixed points on perpendicular axes forms a circle with a specific relationship to those points.
In triangle $OAC$ where $O = (0,0)$, $A = \left(\frac{a}{2}, 0\right)$, $C = \left(0, \frac{b}{2}\right)$, apply the Pythagorean theorem: $OA^2 + AC^2 = OC^2$ gives $\frac{a^2}{4} + \left(h - \frac{a}{2}\right)^2 + \left(k - \frac{b}{2}\right)^2 = h^2 + k^2$. Expanding and simplifying yields $h^2 + k^2 - ah - bk + \frac{a^2 + b^2}{8} = 0$. Therefore, the locus is $x^2 + y^2 - ax - by + \frac{a^2 + b^2}{8} = 0$.
Correct Answer: 3