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The center of the circle with equation \(x^2+y^2=4x+12y-39\)  is the point (h,k) . What is h+k?

tertre  Mar 14, 2017

Best Answer 

 #1
avatar+5555 
+6

Ohhh I can do this one!

 

\(x^2+y^2=4x+12y-39 \\ x^2-4x+y^2-12y=-39 \\ x^2-4x+4+y^2-12y+36=-39+4+36 \\ (x-2)(x-2)+(y-6)(y-6)=1 \\ (x-2)^2 + (y-6)^2 = 1\)

 

So the center is located at (2,6)

 

2 + 6 = 8     :)

hectictar  Mar 14, 2017
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2+0 Answers

 #1
avatar+5555 
+6
Best Answer

Ohhh I can do this one!

 

\(x^2+y^2=4x+12y-39 \\ x^2-4x+y^2-12y=-39 \\ x^2-4x+4+y^2-12y+36=-39+4+36 \\ (x-2)(x-2)+(y-6)(y-6)=1 \\ (x-2)^2 + (y-6)^2 = 1\)

 

So the center is located at (2,6)

 

2 + 6 = 8     :)

hectictar  Mar 14, 2017
 #2
avatar+1227 
+5

Bingo! Nice!

tertre  Mar 14, 2017

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