Answered

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

Sagot :

[tex]x^2+67+y^2=8y+20x[/tex]

Let's rewrite the expression as:

[tex]\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}[/tex]

Which is the standard equation of a circle:

[tex](x-h)^2+(y-k)^2=r^2[/tex]

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

Therefore, the center is:

[tex]\begin{gathered} (h,k)=(10,4) \\ \end{gathered}[/tex]

And the radius is:

[tex]r=\sqrt[]{49}=7[/tex]

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