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Find the center and the radius of the circle $9 x^2 + 9 y^2 - 2 y = 0$

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Answer:

$\left(0, \dfrac{1}{ 9 }\right)$ and $\dfrac {1}{ 9 }$

Step by Step Explanation:

1. The equation of a circle with center at $(a, b)$ and radius $r$ is given by
   $(x - a)^2 + (y - b)^2 = r^2$
2. The given equation is
   $\begin{align} & 9 x^2 + 9 y^2 - 2 y = 0\\ \implies & x^2 + \left(y^2 - \dfrac{ 2 }{ 9 }y \right) = 0 \end{align}$
3. Completing the squares within the parentheses
   $\begin{align} & \implies x^2 + \left(y^2 - \dfrac{ 2 }{ 9 }y + \dfrac{1}{ 81 }\right) = 0 + \dfrac{1}{ 81 } \\ & \implies (x - 0)^2 + \left(y - \dfrac{1}{ 9 }\right)^2 = \dfrac{1}{ 81 } \\ & \implies (x - 0)^2 + \left(y - \dfrac{1}{ 9 }\right)^2 = \left(\dfrac{1}{ 9 }\right)^2 && ...(1) \space\space\space\space\space \end{align}$
4. On comparing eq$(1)$ with the standard form of the equation of the circle, we get,
   $\implies a = 0, b = \dfrac{1}{ 9 },$ and $r = \dfrac {1}{ 9 }$
   Hence, the center of the circle is $\left(0, \dfrac{1}{ 9 }\right)$ and the radius of the circle is $\dfrac {1}{ 9 }$.

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