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What is the general form of the equation of a circle given the Center (-1,2) and Solution Point (0,0)?

1 Answer

Alan P. · mason m

Dec 11, 2015

${(x + 1)}^{2} + {(y - 2)}^{2} = 5$ $(x+1)^2+(y-2)^2=5$

Explanation:

The general form for a circle with center $(a, b)$ $(a,b)$ and radius $r$ $r$ is
$XXX {(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $color(white)("XXX")(x-a)^2+(y-b)^2=r^2$

With center $(- 1, 2)$ $(-1,2)$ and given that $(0, 0)$ $(0,0)$ is a solution (i.e. a point on the circle),
according to the Pythagorean Theorem:
$XXX r^{2} = {(- 1 - 0)}^{2} + {(2 - 0)}^{2} = 5$ $color(white)("XXX")r^2=(-1-0)^2+(2-0)^2 = 5$

and since the center is $(a, b) = (- 1, 2)$ $(a,b)=(-1,2)$

by applying the general formula we get:
$XXX {(x + 1)}^{2} + {(y - 2)}^{2} = 5$ $color(white)("XXX")(x+1)^2+(y-2)^2=5$

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