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The center of the circle with equation \(x^2+y^2=8x-6y-20\)  is the point (x,y) . What is x+y ?

tertre  Mar 14, 2017

Best Answer 

 #1
avatar+4728 
+6

\(x^2+y^2=8x-6y-20 \\ x^2-8x+y^2+6y=-20 \\ x^2-8x+16+y^2+6y+9=-20+16+9 \\ (x-4)(x-4)+(y+3)(y+3)=5 \\ (x-4)^2 + (y+3)^2 = 5\)

 

So the center is located at (4,-3)

4 + (-3) = 4 - 3 = 1

hectictar  Mar 14, 2017
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2+0 Answers

 #1
avatar+4728 
+6
Best Answer

\(x^2+y^2=8x-6y-20 \\ x^2-8x+y^2+6y=-20 \\ x^2-8x+16+y^2+6y+9=-20+16+9 \\ (x-4)(x-4)+(y+3)(y+3)=5 \\ (x-4)^2 + (y+3)^2 = 5\)

 

So the center is located at (4,-3)

4 + (-3) = 4 - 3 = 1

hectictar  Mar 14, 2017
 #2
avatar+1167 
0

Bingo!

tertre  Mar 14, 2017

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