SOLUTION: Graph x^2 -2x +y^2=8
y^2=-x^2 -2x+8
y=-x^2 -2x+8 (this line + or - the Square root of the entire equation)
I dont know if I am entering it wrong into the calc., but it is suppos
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-> SOLUTION: Graph x^2 -2x +y^2=8
y^2=-x^2 -2x+8
y=-x^2 -2x+8 (this line + or - the Square root of the entire equation)
I dont know if I am entering it wrong into the calc., but it is suppos
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Question 261817: Graph x^2 -2x +y^2=8
y^2=-x^2 -2x+8
y=-x^2 -2x+8 (this line + or - the Square root of the entire equation)
I dont know if I am entering it wrong into the calc., but it is suppose to be a graph of a circle. Answer by solver91311(24713) (Show Source):
Divide the coefficient on the first degree term by 2 and square the result. -2 divided by 2 is -1. -1 squared is 1. Add the result to both sides of the equation:
Factor the perfect square trinomial in :
Re-write as :
A circle centered at and with radius has an equation:
So find the point and draw a circle with radius 3 centered there.
