The length of the common chord of the circles | vedclass

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(A) The equations of the circles are:$S_1: (x - a)^2 + (y - b)^2 = c^2$$S_2: (x - b)^2 + (y - a)^2 = c^2$Subtracting $S_2$ from $S_1$ gives the equati

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$\sqrt{4c^2 - 2(a - b)^2}$

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(A) The equations of the circles are:$S_1: (x - a)^2 + (y - b)^2 = c^2$$S_2: (x - b)^2 + (y - a)^2 = c^2$Subtracting $S_2$ from $S_1$ gives the equation of the common chord:$(x - a)^2 - (x - b)^2 + (y - b)^2 - (y - a)^2 = 0$$(x^2 - 2ax + a^2) - (x^2 - 2bx + b^2) + (y^2 - 2by + b^2) - (y^2 - 2ay + a^2) = 0$$-2ax + 2bx - 2by + 2ay = 0$$-2a(x - y) + 2b(x - y) = 0$$(2b - 2a)(x - y) = 0$Since $a 
eq b$,we have $x - y = 0$,or $y = x$.Substitute $y = x$ into the first circle equation:$(x - a)^2 + (x - b)^2 = c^2$$x^2 - 2ax + a^2 + x^2 - 2bx + b^2 = c^2$$2x^2 - 2(a + b)x + (a^2 + b^2 - c^2) = 0$The distance $d$ from the center $(a, b)$ to the line $x - y = 0$ is:$d = \frac{|a - b|}{\sqrt{1^2 + (-1)^2}} = \frac{|a - b|}{\sqrt{2}}$The length of the common chord is $2\sqrt{r^2 - d^2}$:$= 2\sqrt{c^2 - \frac{(a - b)^2}{2}}$$= 2\sqrt{\frac{2c^2 - (a - b)^2}{2}}$$= \sqrt{2(2c^2 - (a - b)^2)}$$= \sqrt{4c^2 - 2(a - b)^2}$

The length of the common chord of the circles $(x - a)^2 + (y - b)^2 = c^2$ and $(x - b)^2 + (y - a)^2 = c^2$ is

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