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Answer:
The center would be (-8,12).
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Step-by-step explanation:
First,we're going to have to simplify the equation. Here is the work below:
[tex]x^2+y^2+16x-24y+159=0\\\\(x+8)^2-64+y^2-24y+159=0\\\\(x+8)^2+y^2-24y=-159+64\\\\(x+8)^2+(y-12)^2-144=-159+64\\\\(x+8)^2+(y-12)^2=144-159+64\\\\(x+8)^2+(y-12)^2=49\\\\[/tex]
Now, you must use the "form of a circle" equation in order to find out the center and radius of a circle.
The equation is:
[tex](x-h)^2+(y-k)^2=r^2\\\\r=7\\h=-8\\k=12[/tex]
Since we got the answer of:
[tex](x+8)^2+(y-12)^2=49[/tex]
It matches with our equation of [tex](x-h)^2+(y-k)^2=r^2[/tex]
Our "h" represents the x axis on the graph, our "k" represents our y axis on a graph.
Therefore our x value would be -8 and our y value would be 12.
Your FINAL answer should be (-8,12)
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