Question

17

What is the center and radius of the circle with the equation
2? + y2 - 2x + 4y - 4 = 0?
1) Radius:
Center:

Answer 1

`Center = \boxed{1}, \: \boxed{ - 2} \: ; \: Radius = \boxed{3}`

Explanation:

`{x}^(2) + {y}^(2) - 2x + 4y - 4 = 0 \\ \\ ({x}^(2) - 2x + 1 - 1) + ( {y}^(2) + 4y + 4 - 4) - 4 = 0 \\ \\ ( {x}^(2) - 2x + 1) + ( {y}^(2) + 4y + 4) - 9 = 0 \\ \\ {(x - 1)}^(2) + {(y + 2)}^(2) = 9 \\ \\ {(x - 1)}^(2) + {(y + 2)}^(2) = {3}^(2) \\ \\ equating \: it \: with \\ \\ {(x - h)}^(2) + {(y - k)}^(2) = {r}^(2) \\ \\ h = 1 \\ k = - 2 \\ r = 3 \\ \\ center = \boxed{1} \: \boxed{ - 2} \: \: radius = \boxed{3}`