# How do you find the center and radius for 21y^2+19x^2+83=4x−2x^2−84y?

Aug 22, 2016

The center of circle is $\left(\frac{2}{21} , - 2\right)$ and radius is $\frac{5}{21}$

#### Explanation:

The equation of a circle is of the type ${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$, i.e. coefficients of $x$ and $y$ are equal and that of $x y$ is $0$. In such a case coordinates of center are $\left(- g , - f\right)$ and radius is given by $\sqrt{{g}^{2} + {f}^{2} - c}$.

Here equation is given to be $21 {y}^{2} + 19 {x}^{2} + 83 = 4 x - 2 {x}^{2} - 84 y$ or

$21 {x}^{2} + 21 {y}^{2} - 4 x + 84 y + 83 = 0$ or

${x}^{2} + {y}^{2} - \frac{4}{21} x + 4 y + \frac{83}{21} = 0$.

Hence center is $\left(\frac{2}{21} , - 2\right)$ and radius is

$\sqrt{{\left(\frac{2}{21}\right)}^{2} + {2}^{2} - \frac{83}{21}}$

= $\sqrt{\frac{4}{441} + \frac{84 - 83}{21}}$

= $\sqrt{\frac{4}{441} + \frac{1}{21}}$

= $\sqrt{\frac{4}{441} + \frac{21}{441}}$

= $\sqrt{\frac{25}{441}} = \frac{5}{21}$

Hence the center of circle is $\left(\frac{2}{21} , - 2\right)$ and radius is $\frac{5}{21}$