Circle has the equation x^2+y^2+2x-2y-14=0, how do you graph the circle using the center (h,k) radius r?

1 Answer
Dec 5, 2015

I found a circle centred at $\left(- 1 , 1\right)$ and with radius $r = 4$.

Explanation:

I would write:
$\left({x}^{2} + 2 x\right) + \left({y}^{2} - 2 y\right) = 14$
add and subtract some values:
$\left({x}^{2} + 2 x \textcolor{red}{+ 1 - 1}\right) + \left({y}^{2} - 2 y \textcolor{red}{+ 1 - 1}\right) = 14$
$\left({x}^{2} + 2 x + 1\right) + \left({y}^{2} - 2 x + 1\right) - 2 = 14$
${\left(x + 1\right)}^{2} + {\left(y - 1\right)}^{2} = 16$
or
${\left(x - \left(- 1\right)\right)}^{2} + {\left(y - \left(1\right)\right)}^{2} = 16$
${\left(x - \left(- 1\right)\right)}^{2} + {\left(y - \left(1\right)\right)}^{2} = {4}^{2}$
Which is in the form:
${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$
giving you a circle centred at $\left(- 1 , 1\right)$ and with radius $r = 4$.