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What is the Center and radius of x2+67+y2=8y+20x

User Eric Bynum
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5 votes

x^2+67+y^2=8y+20x

Let's rewrite the expression as:


\begin{gathered} x^2+67+y^2-8y-20x=0 \\ so\colon \\ (x-10)^2+(y-4)^2-49=0 \\ (x-10)^2+(y-4)^2=49 \end{gathered}

Which is the standard equation of a circle:


(x-h)^2+(y-k)^2=r^2

Where (h,k) is the coordinates of center of the circle and r is the radius

Therefore, the center is:


\begin{gathered} (h,k)=(10,4) \\ \end{gathered}

And the radius is:


r=\sqrt[]{49}=7

What is the Center and radius of x2+67+y2=8y+20x-example-1
User Kugutsumen
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