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

Algebra ->  Test -> 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      Log On


   

Question 1085947: 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.
x²+y²-8x+6y+40=0
Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
.
The standard procedure for such problems is completing the squares:


x%5E2+%2B+y%5E2+-+8x+%2B+6y = -40,

%28x-4%29%5E2+%2B+%28y%2B3%29%5E2 = -40+-+16+-+9.


Stop here.


You see that the left side is the sum of two squares, while the right side is the negative real number.


It implies that neither last equation, nor equivalent original equation has a  solution.


So, the set of solutions is empty. It is the answer to your problem.