# How do you find the standard form of x^2 + y^2 + 8x + 2y - 8 = 0 and what kind of a conic is it?

Dec 7, 2015

Shape: Circle
${\left(x + 4\right)}^{2} + {\left(y + 1\right)}^{2} = 9$

#### Explanation:

Remember: Some of the formula for conic are:
Circle: ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$
Ellipse: ${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$
Hyperbola:${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$

Step 1: Complete the so to determine the form

${x}^{2} + 8 x + {y}^{2} + 2 y = 8$

(x^2 + 8x+color(red)(16)) + (y^2 + 2y+color(red)(1)) = -8+ color(red)(16+ 1

*Note: to complete the square, color(red)(c= (b/2)^2
${\left(x + 4\right)}^{2} + {\left(y + 1\right)}^{2} = 9$

Center: $\left(- 4 , - 1\right)$