Answer to Question #259694 in Calculus for bbz

Question

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

Answer 1

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

if

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

then we get

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