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Find the center and the radius of the circle $5 x^2 + 5 y^2 - 2 x = 0$

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Answer:

$\left(\dfrac{1}{ 5 }, 0\right)$ and $\dfrac {1}{ 5 }$

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
   $\begin{align} & 5 x^2 + 5 y^2 - 2 x = 0\\ \implies & x^2 + y^2 - \dfrac{ 2 }{ 5 }x = 0 \\ \implies & \left(x^2 - \dfrac{ 2 }{ 5 }x \right) + y^2 = 0 \end{align}$
3. Completing the squares within the parentheses
   $\begin{align} & \implies \left(x^2 - \dfrac{ 2 }{ 5 }x + \dfrac{1}{ 25 }\right) + y^2 = 0 + \dfrac{1}{ 25 } \\ & \implies \left(x - \dfrac{1}{ 5 }\right)^2 + (y - 0)^2 = \dfrac{1}{ 25 } \\ & \implies \left(x - \dfrac{1}{ 5 }\right)^2 + (y - 0)^2 = \left(\dfrac{1}{ 5 }\right)^2 \space\space\space\space\space ...(1) \end{align}$
4. On comparing eq$(1)$ with the standard form of the equation of the circle, we get,
   $\implies a = \dfrac{1}{ 5 }, b = 0,$ and $r = \dfrac {1}{ 5 }$
   Hence, the center of the circle is $\left(\dfrac{1}{ 5 }, 0\right)$ and the radius of the circle is $\dfrac {1}{ 5 }$.

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