SOLUTION: Equation of a circle if I have P (-5,2) and Q (3,-8) I have it figured as -1, -3
then r= square root of (-1-3)^2 + (-3+8)^2 = sqaure root 41 so the equation of the circle would be
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-> SOLUTION: Equation of a circle if I have P (-5,2) and Q (3,-8) I have it figured as -1, -3
then r= square root of (-1-3)^2 + (-3+8)^2 = sqaure root 41 so the equation of the circle would be
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Question 385741: Equation of a circle if I have P (-5,2) and Q (3,-8) I have it figured as -1, -3
then r= square root of (-1-3)^2 + (-3+8)^2 = sqaure root 41 so the equation of the circle would be (x+1)^2 + (y+3)^2 = 41.
But thing seems a miss.... could you look this over for me? Thanks
Melody Answer by robertb(5830) (Show Source):
You can put this solution on YOUR website! You're saying that (-1,-3) is the center of the circle. Are the points P and Q given as the endpoints of a diameter? Incidentally if (-1,-3) is indeed the center of the circle then 41 is the square of the radius...
