Math Problem Solver

The center of the circle $9 x^{2} - 12 x + 9 y^{2} + 24 y - 16 = 0$ is at $\left( \frac{2}{3}, \ - \frac{4}{3}\right)$.

The standard form of a circle is $(x-h)^2+(y-k)^2=r^2$ where $(h, k)$ is the center and $r$ is the radius. Convert the equation into standard form by completing the square for both $x$-dependent terms and $y$-dependent terms.$$\begin{aligned}9 x^{2} - 12 x + 9 y^{2} + 24 y - 16 &= 0\\9 x^{2} - 12 x + 9 y^{2} + 24 y &= 16\\9 \left(x^{2} - \frac{4 x}{3}\right) + 9 \left(y^{2} + \frac{8 y}{3}\right) &= 16\\9 \left(x^{2} - \frac{4 x}{3} + \frac{4}{9}\right) + 9 \left(y^{2} + \frac{8 y}{3} + \frac{16}{9}\right) &= 16 + 9 \cdot \frac{4}{9} + 9 \cdot \frac{16}{9}\\9 \left(x - \frac{2}{3}\right)^{2} + 9 \left(y + \frac{4}{3}\right)^{2} &= 4 + 16 + 16\\9 \left(x - \frac{2}{3}\right)^{2} + 9 \left(y + \frac{4}{3}\right)^{2} &= 36\\\left(x - \frac{2}{3}\right)^{2} + \left(y + \frac{4}{3}\right)^{2} &= \frac{36}{9}\\\left(x - \frac{2}{3}\right)^{2} + \left(y + \frac{4}{3}\right)^{2} &= 4\end{aligned}$$The center is at $(h, k)$ or $\left( \frac{2}{3}, \ - \frac{4}{3}\right)$.

The radius of the circle $9 x^{2} - 12 x + 9 y^{2} + 24 y - 16 = 0$ is $2$.

The standard form of a circle is $(x-h)^2+(y-k)^2=r^2$ where $(h, k)$ is the center and $r$ is the radius. Convert the equation into standard form by completing the square for both $x$-dependent terms and $y$-dependent terms.$$\begin{aligned}9 x^{2} - 12 x + 9 y^{2} + 24 y - 16 &= 0\\9 x^{2} - 12 x + 9 y^{2} + 24 y &= 16\\9 \left(x^{2} - \frac{4 x}{3}\right) + 9 \left(y^{2} + \frac{8 y}{3}\right) &= 16\\9 \left(x^{2} - \frac{4 x}{3} + \frac{4}{9}\right) + 9 \left(y^{2} + \frac{8 y}{3} + \frac{16}{9}\right) &= 16 + 9 \cdot \frac{4}{9} + 9 \cdot \frac{16}{9}\\9 \left(x - \frac{2}{3}\right)^{2} + 9 \left(y + \frac{4}{3}\right)^{2} &= 4 + 16 + 16\\9 \left(x - \frac{2}{3}\right)^{2} + 9 \left(y + \frac{4}{3}\right)^{2} &= 36\\\left(x - \frac{2}{3}\right)^{2} + \left(y + \frac{4}{3}\right)^{2} &= \frac{36}{9}\\\left(x - \frac{2}{3}\right)^{2} + \left(y + \frac{4}{3}\right)^{2} &= 4\end{aligned}$$The radius is $r$, or $\sqrt{4}$. $\sqrt{4} = \pm 2$, and a radius is always positive, so the radius is $2$.