so my equation is x^{2}+y^{2}-4x+2y-4=0
i know the formula is (x-h)^{2}+(y-k)^{2}=r^{2}
x^{2}-4x+y^{2}+2y-4=0
the Y is (y-2)^{2} which makes my k equal to positive 2 just having trouble with the h and radius
so my equation is x^{2}+y^{2}-4x+2y-4=0
i know the formula is (x-h)^{2}+(y-k)^{2}=r^{2}
x^{2}-4x+y^{2}+2y-4=0
the Y is (y-2)^{2} which makes my k equal to positive 2 just having trouble with the h and radius
Hello, M670!
Sorry, your algebra is off . . .
$\displaystyle \text{Find the center and radius of the circle: }\:x^2+y^2-4x+2y-4\:=\:0$
We must "complete the square".
$\displaystyle \begin{array}{c}x^2 + y^2 - 4x + 2y - 4 \:=\:0 \\ \\ x^2 - 4x + y^2 + 2y \:=\:4 \\ \\ x^2 - 4x \,{\color{red}+\, 4} + y^2 + 2y {\color{blue}\,+\, 1} \:=\:4{\color{red}\,+\,4}{\color{blue}\,+\,1} \\ \\ (x-2)^2 +(y+1)^2 \:=\:9 \end{array}$
The circle has center $\displaystyle (2,\text{-}1)$ and radius $\displaystyle r = 3.$
ok I understand that question but given this example here 5 x^2 - 10 x +5 y^2 - 8 y - 6 = 0 .
I worked it down to (x-1)^2 and (y-0.8)^2 =
I divided out 5 and was left with x^2-2x+y^2-1.6y-1.2=0
then its x^2-2x+1 + y^2-1.6y ?= 1.2-1 (?)