SOLUTION: Question: Identify the radius and center, if the graph of the equation is a circle then sketch it. Otherwise, specify if the graph is a single point (and what this point is), or ha

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Question 1085764: Question: Identify the radius and center, if the graph of the equation is a circle then sketch it. Otherwise, specify if the graph is a single point (and what this point is), or has no point at all.
1.8x²+8y²-28+12y+221=0
2.x²+y²-8x+6y+40=0
3.x²+y²+22x+18y+215=0
4.x²+y²-18x+12y+144=0
Answer by Fombitz(32388)   (Show Source): You can put this solution on YOUR website!
Complete the square to get the equation into the general equation of a circle.

I'll do one, you do the others in the same way.
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I'll assume you made a mistake and that it's and not .






Since the radius term is negative, there is no point at all.
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If it equals zero, then the center is the single point.
If its positive, then the square root of the value is the radius of the circle.
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Do the others the same way.
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