Equation to the circle whose one of the diameters is the common chord of $(\mathrm{x}-\mathrm{a})^{2}+\mathrm{y}^{2}=\mathrm{a}^{2},\ \mathrm{x}^{2}+(\mathrm{y}-\mathrm{b})^{2}=\mathrm{b}^{2}$ is
- $(\mathrm{a}^{2}+\mathrm{b}^{2})(\mathrm{x}^{2}+\mathrm{y}^{2})=2\mathrm{a}\mathrm{b}(\mathrm{b}\mathrm{x}+\mathrm{a}\mathrm{y})$
- $(\mathrm{a}^{2}+\mathrm{b}^{2})(\mathrm{x}^{2}+\mathrm{y}^{2})=2\mathrm{a}\mathrm{b}(\mathrm{a}\mathrm{x}+\mathrm{b}\mathrm{y})$
- $\mathrm{x}^{2}+\mathrm{y}^{2}=\dfrac {2\mathrm{a}\mathrm{b}}{(\mathrm{a}^{2}+\mathrm{b}^{2}) (ax-by)}$
- $\mathrm{x}^{2}+\mathrm{y}^{2}= \dfrac {ab}{(\mathrm{a}^{2}+\mathrm{b}^{2})(\mathrm{a}\mathrm{x}+\mathrm{b}\mathrm{y})}$
The common chord of the two circles is found by subtracting the equations, resulting in ax + by = 0. The equation of a circle passing through the intersection of two circles is S1 + k(S2-S1) = 0, or more simply, the family of circles passing through the intersection. Using the condition that the common chord is a diameter, we derive the correct equation.
Expanding the given equations gives x squared plus y squared minus 2ax equals 0 and x squared plus y squared minus 2by equals 0. The equation of their common chord is bx minus ay equals 0, which passes through the origin. The required circle having this chord as its diameter must pass through the origin and the intersections of the two given circles, making its equation x squared plus y squared plus 2gx plus 2fy equals 0. By solving the system for the common passing points and substituting back to find the linear coefficients, the resulting equation simplifies to (a squared + b squared)(x squared + y squared) = 2ab(bx + ay).