Two circles centered at (2,3) and (5,6) inter | vedclass

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(C) Let the radius of both circles be $r$. The equation of the circle with center at $(2,3)$ is: $S_{1} \equiv (x-2)^{2} + (y-3)^{2} = r^{2} \quad \do

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$x+y-8=0$

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(C) Let the radius of both circles be $r$. The equation of the circle with center at $(2,3)$ is: $S_{1} \equiv (x-2)^{2} + (y-3)^{2} = r^{2} \quad \dots(i)$ The equation of the circle with center at $(5,6)$ is: $S_{2} \equiv (x-5)^{2} + (y-6)^{2} = r^{2} \quad \dots(ii)$ The equation of the common chord is given by the radical axis $S_{1} - S_{2} = 0$. $(x-2)^{2} + (y-3)^{2} - ((x-5)^{2} + (y-6)^{2}) = 0$ $(x^{2} - 4x + 4 + y^{2} - 6y + 9) - (x^{2} - 10x + 25 + y^{2} - 12y + 36) = 0$ $(x^{2} + y^{2} - 4x - 6y + 13) - (x^{2} + y^{2} - 10x - 12y + 61) = 0$ $6x + 6y - 48 = 0$ Dividing by $6$,we get: $x + y - 8 = 0$

Two circles centered at $(2,3)$ and $(5,6)$ intersect each other. If the radii are equal,the equation of the common chord is

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