Question 25752
diameter that has endpoints (0,-7) and (4,-10) 
EQN OF A CIRCLE WITH DIAMETER AT END POINTS (X1,Y1) AND (X2-Y2) IS GIVEN BY 
{(Y-Y2)/(X-X2)}*{(Y-Y1)/(X-X1)=-1...OR
(Y-Y2)*(Y-Y1)+(X-X2)*(X-X1)=0..SO WE GET
(Y+7)(Y+10)+(X)(X-4)=0
Y^2+17y+70+X^2-4X=0
(Y+17/2)^2-(17/2)^2+(X-2)^2-4+70=0
(X-2)^2+(Y+17/2)^2=(17/2)^2-66=6.25
VERY GOOD YOU GOT EXACTLY CORRECT ANSWER FOR H=2 AND K=-8.5...GOOD JOB...KEEP IT UP...
R^2 IS GIVEN BY 6.25..OK