UAE
Find the center and the radius of the circle ^@9 x^2 + 9 y^2 - 2 y = 0^@
Answer:
^@\left(0, \dfrac{1}{ 9 }\right)^@ and ^@\dfrac {1}{ 9 }^@
- 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} & 9 x^2 + 9 y^2 - 2 y = 0\\ \implies & x^2 + \left(y^2 - \dfrac{ 2 }{ 9 }y \right) = 0 \end{align}^@ - 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}^@ - 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 }^@.
