The length of the common chord of the two cir | vedclass

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(B) The equations of the circles are:$S_1: x^2+y^2-2ax=0$$S_2: x^2+y^2-2by=0$The equation of the common chord is given by $S_1 - S_2 = 0$:$(x^2+y^2-2a

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

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(B) The equations of the circles are:$S_1: x^2+y^2-2ax=0$$S_2: x^2+y^2-2by=0$The equation of the common chord is given by $S_1 - S_2 = 0$:$(x^2+y^2-2ax) - (x^2+y^2-2by) = 0$$-2ax + 2by = 0$$ax - by = 0$The centre of circle $S_1$ is $C_1(a, 0)$ and its radius is $r_1 = a$.The perpendicular distance $d$ from the centre $C_1(a, 0)$ to the common chord $ax - by = 0$ is:$d = \frac{|a(a) - b(0)|}{\sqrt{a^2 + (-b)^2}} = \frac{a^2}{\sqrt{a^2+b^2}}$The length of the common chord is $2\sqrt{r_1^2 - d^2}$:$= 2\sqrt{a^2 - \left(\frac{a^2}{\sqrt{a^2+b^2}}\right)^2}$$= 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 two circles $(x-a)^2+y^2=a^2$ and $x^2+(y-b)^2=b^2$ is

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