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(x−A)2+(y −B)2 = (x−C)2+(y−D)2\left(x-A\right)^2+\left(y\:-B\right)^2\:=\:\left(x-C\right)^2+\left(y-D\right)^2

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Final answer to the exercise

y=−2xc+c2+d2+2xa−a2−b22(−b+d)y=\frac{-2xc+c^2+d^2+2xa-a^2-b^2}{2\left(-b+d\right)}

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(x−A)2+(y −B)2 = (x−C)2+(y−D)2
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