# How do you write the equation x^2+y^2+8x-10y-1=0 in standard form and find the center and radius?

Feb 11, 2018

Standard form of equation of a circle is ${\left(x + 4\right)}^{2} + {\left(y - 5\right)}^{2} = {\left(\sqrt{42}\right)}^{2}$. Center is $\left(- 4 , 5\right)$ and radius is $\sqrt{42}$.

#### Explanation:

${x}^{2} + {y}^{2} + 8 x - 10 y - 1 = 0$ can be written as

${x}^{2} + 8 x + {y}^{2} - 10 y = 1$

or $\left({x}^{2} + 8 x + 16\right) + \left({y}^{2} - 10 y + 25\right) = 1 + 16 + 25$

or ${\left(x + 4\right)}^{2} + {\left(y - 5\right)}^{2} = 42$

or ${\left(x + 4\right)}^{2} + {\left(y - 5\right)}^{2} = {\left(\sqrt{42}\right)}^{2}$

This is the standard form of equation of a circle and

indicates that the point $\left(x , y\right)$ moves so that its distance from the point $\left(- 4 , 5\right)$ is always $\sqrt{42}$

Hence, center of circle is $\left(- 4 , 5\right)$ and radius is $\sqrt{42}$

graph{x^2+y^2+8x-10y-1=0 [-18, 10, -2, 12]}