Question 36510: (1) find the center of the circle whose equation is
x^2+ y^2-3x+8y-8=0
(2) write the equation of the circle with center (-2,1) and which passes through (3,4).
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! STD.EQN. OF A CIRCLE IS
(X-H)^2+(Y-K)^2=R^2...
WHERE (H,K)IS THE CENTRE AND R IS THE RADIUS.USING THIS FORMULA...
(1) find the center of the circle whose equation is
x^2+ y^2-3x+8y-8=0
{X^2-2*(3/2)X+(3/2)^2}+{Y^2+2*4*Y+4^2}-(3/2)^2-4^2-8=0
{X-(3/2)}^2+{Y+4}^2=24+9/4=105/4..HENCE CENTRE OF CIRCLE IS {(3/2),-4}
(2) write the equation of the circle with center (-2,1) and which passes through (3,4).
EQN IS (X+2)^2+(Y-1)^2=R^2...IT IS` PASSING THROUGH (3,4)..SO...
(3+2)^2+(4-1)^2=R^2
25+9=34=R^2.HENCE EQN.OF CIRCLE IS
(X+2)^2+(Y-1)^2=34
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