SOLUTION: (x-1)^2 + (y-8)^2 = 16 I've tried giving x a value and solving for y. Here is what I tried... (1-1)^2 + (y-8)^2 = 16 0 + (y-8)^2 = 16 (y-8)(y-8) = 16 y^2 -8y -8y =64 = 16 y^2

Algebra ->  Graphs -> SOLUTION: (x-1)^2 + (y-8)^2 = 16 I've tried giving x a value and solving for y. Here is what I tried... (1-1)^2 + (y-8)^2 = 16 0 + (y-8)^2 = 16 (y-8)(y-8) = 16 y^2 -8y -8y =64 = 16 y^2      Log On


   

Question 72487: (x-1)^2 + (y-8)^2 = 16
I've tried giving x a value and solving for y. Here is what I tried...
(1-1)^2 + (y-8)^2 = 16
0 + (y-8)^2 = 16
(y-8)(y-8) = 16
y^2 -8y -8y =64 = 16
y^2 - 16y + 64 = 16
y^2 - 8(y-y+8) = 16
y^2 + 8 =2
y^2 = 16
y = 4
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
%28x-1%29%5E2+%2B+%28y-8%29%5E2+=+16
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Here's a point to the right direction. Look in your textbook for the standard form of the
equation of a circle.
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The above problem is a circle whose center is found by getting its x and y values through solving
the two equations x-1 = 0 and y - 8 = 0.
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The radius of the circle is the square root of the right side.
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In summary, the graph of this equation is a circle whose center is located located at (1,8) and
whose radius is 4.
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Hope this points you in a little better direction. If you study how the equation for
a circle is derived you will get a little better idea of how the above form comes about.
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And don't feel too badly about this. The only reason I saw the answer immediately
was because the equation was written in a form that I recognized from a long time ago.