The length of the common chord of the circles | vedclass

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(C) The equations of the circles are $S_1: x^2 - 2ax + a^2 + y^2 = a^2 \implies x^2 + y^2 - 2ax = 0$ and $S_2: x^2 + y^2 - 2by + b^2 = b^2 \implies x^

Vedclass · Accepted Answer

$\frac{2ab}{\sqrt{a^2 + b^2}}$

Vedclass · Answer

(C) The equations of the circles are $S_1: x^2 - 2ax + a^2 + y^2 = a^2 \implies x^2 + y^2 - 2ax = 0$ and $S_2: x^2 + y^2 - 2by + b^2 = b^2 \implies x^2 + y^2 - 2by = 0$.The equation of the common chord is given by $S_1 - S_2 = 0$,which is $-2ax + 2by = 0$,or $ax - by = 0$.The center of the first circle is $(a, 0)$ and its radius is $r_1 = a$.The perpendicular distance $p$ from the center $(a, 0)$ to the line $ax - by = 0$ is $p = \frac{|a(a) - b(0)|}{\sqrt{a^2 + (-b)^2}} = \frac{a^2}{\sqrt{a^2 + b^2}}$.The length of the common chord is $2\sqrt{r_1^2 - p^2} = 2\sqrt{a^2 - \frac{a^4}{a^2 + b^2}}$.$= 2\sqrt{\frac{a^2(a^2 + b^2) - a^4}{a^2 + b^2}} = 2\sqrt{\frac{a^4 + a^2b^2 - a^4}{a^2 + b^2}} = 2\sqrt{\frac{a^2b^2}{a^2 + b^2}} = \frac{2ab}{\sqrt{a^2 + b^2}}$.

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

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