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Read the equation in standard form for the circle passing through (-3,0) centered at the origin.

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the standard equation of circles is:


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

in which the circle is centered at (0,0).

Since the circle is centered at the origin (h,k) is (0,0), then, the equation can be reduced to:


x^2+y^2=r^2

then, since the circle passes through the point (-3,0) the radius is equal to 3,

finally, the standard equation can be replaced with the radius


\begin{gathered} x^2+y^2=3^2 \\ x^2+y^2=9 \end{gathered}

Answer:

The equation for a circle centered at the origin that passes through the point (-3,0) is:


x^2+y^2=9

User Chachmu
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