Answer in Calculus for bbz #259694

Question #259694

why did (7-a)^2 + (2-b)^2 = R^2 become (2-b)^2 - b^2= 0

Expert's answer

$(2-b)^2 + (7-a)^2-R^2=0$

$b^2=R^2- (7-a)^2$

then we get

$(2-b)^2 = b^2$

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