Question 25752: find the standard form for the equation of a circle (x-h)sqrd + (y-k)sqrd=rsqrd with a diameter that has endpoints (0,-7) and (4,-10) . i gor h=2 and k=-8.5 but rsqrd=?
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! 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
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