What is the center and radius of the circle x^2+y^2-2x+4y-4=0?

Mar 10, 2017

centre$\text{ } \left(1 , - 2\right)$

radius$\text{ } 3$

Explanation:

The standard eqn of a circle with centre $\left(a , b\right)$ and radius $r$

is: ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

for $\text{ "x^2+y^2-2x+4y-4=0" }$we need to complete the square.

$\text{ "x^2+y^2-2x+4y-4=0" }$

$\textcolor{red}{{x}^{2} - 2 x} + \textcolor{b l u e}{{y}^{2} + 4 y} - 4 = 0$

$\textcolor{red}{\left({x}^{2} - 2 x + {1}^{2}\right)} + \textcolor{b l u e}{\left({y}^{2} + 4 y + {2}^{2}\right)} - {1}^{2} - {2}^{2} - 4 = 0$

$\textcolor{red}{{\left(x - 1\right)}^{2}} + \textcolor{b l u e}{{\left(y + 2\right)}^{2}} - 9 = 0$

${\left(x - 1\right)}^{2} + {\left(y + 2\right)}^{2} = 9$

centre$\text{ } \left(1 , - 2\right)$

radius$\text{ } \sqrt{9} = 3$