What is the center and radius of the circle with equation ${(x - 6)}^{2} + y^{2} = 49$ $(x - 6)^2 + y^2 = 49$ ?

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Answer 1 · 2016-06-03T00:58:57

Center: $(6, 0)$ $(6, 0)$
Radius: $7$ $7$

A circle centered at $(x_{0}, y_{0})$ $(x_0,y_0)$ with radius $r$ $r$ has the equation

${(x - x_{0})}_{0}^{2} + {(y - y_{0})}_{0}^{2} = r^{2}$ $(x-x_0)^2+(y-y_0)^2=r^2$

We can make the given equation fit this form with some slight changes:

${(x - 6)}^{2} + y^{2} = 49$ $(x-6)^2+y^2 = 49$

$\Rightarrow {(x - 6)}^{2} + {(y - 0)}^{2} = 7^{2}$ $=> (x-6)^2+(y-0)^2 = 7^2$

Thus the circle is centered at $(6, 0)$ $(6,0)$ and has radius $7$ $7$

What is the center and radius of the circle with equation (x−6)2+y2=49(x - 6)^2 + y^2 = 49?