Question

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?

Answer 1

I found a circle centred at `(-1,1)` and with radius `r=4`.

Explanation:

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