Find the center and the radius of the circle ^@5 x^2 + 5 y^2 - 2 x = 0^@
Answer:
^@\left(\dfrac{1}{ 5 }, 0\right)^@ and ^@\dfrac {1}{ 5 }^@
- The equation of a circle with center at ^@(a, b)^@ and radius ^@r^@ is given by
^@(x - a)^2 + (y - b)^2 = r^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}^@ - 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}^@ - 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 }^@.
