SOLUTION: What is the center of the curve x^2+y^2-2x-4y-31=0.Please explain.

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Question 1115789: What is the center of the curve x^2+y^2-2x-4y-31=0.Please explain.
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

What is the center of the curve
+x%5E2%2By%5E2-2x-4y-31=0
%28x%5E2-2x%29%2B%28y%5E2-4y%29=31....complete square
%28x%5E2-2x%2Bb%5E2%29-b%5E2%2B%28y%5E2-4y%2Bb%5E2%29+-b%5E2+=31
recal
%28a%2Bb%29%5E2=a%5E2%2B2ab%2Bb%5E2
for x part, you have a=1 and 2ab=-2=>2%2A1%2Ab=-2=>b=-1
for y part, you have a=1 and 2ab=-4=>2%2A1%2Ab=-4=>b=-2
so, you have
%28x%5E2-2x%2B%28-1%29%5E2%29-%28-1%29%5E2%2B%28y%5E2-4y%2B%28-2%29%5E2%29+-%28-2%29%5E2+=31
%28x-1%29%5E2-1%2B%28y-2%29%5E2+-%28-2%29%5E2+=31
%28x-1%29%5E2-1%2B%28y-2%29%5E2+-4+=31
%28x-1%29%5E2%2B%28y-2%29%5E2++=31%2B1%2B4
%28x-1%29%5E2%2B%28y-2%29%5E2++=36
%28x-1%29%5E2%2B%28y-2%29%5E2++=6%5E2
so, you have a circle with h=1, k=2, and r=6
the center is at (1, 2)