Question

A circle passing through origin cuts the coordinate axes at \(A\) and \(B\). If the straight line \(AB\) passes through a fixed point \((x_1,y_1)\), then the locus of the centre of the circle is?

• A. \(\frac{x_1}{x} + \frac{y_1}{y} = 1\)
• B. \(xy = x_1 y_1\)
• C. \(x y_1 + y x_1 = 2\)
• D. \(\frac{x_1}{x} + \frac{y_1}{y} = 2\)

Hint:

Express circle in intercept form and use fixed point condition to find locus of centers.

Answer 1

The Correct Option is D

Equation of circle passing origin and cutting axes at \(A(a,0)\), \(B(0,b)\): \[ \frac{x}{a} + \frac{y}{b} = 1. \] Center of circle: \(\left(\frac{a}{2}, \frac{b}{2}\right)\). Since line \(AB\) passes through \((x_1,y_1)\), \[ \frac{x_1}{a} + \frac{y_1}{b} = 1. \] Substitute \(a = 2h, b=2k\): \[ \frac{x_1}{2h} + \frac{y_1}{2k} = 1 \implies \frac{x_1}{h} + \frac{y_1}{k} = 2. \] Hence locus of center \((h,k)\) is \[ \frac{x_1}{x} + \frac{y_1}{y} = 2. \]