How do you write the equation for a circle with center at (-1,3) passes through the point (-1,4)?

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How do you write the equation for a circle with center at (-1,3) passes through the point (-1,4)?

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Answer 1 · 2016-04-22T09:59:44

${(x + 1)}^{2} + {(y - 3)}^{2} = 1$ $(x+1)^2+(y-3)^2=1$

Explanation:

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$O = (- 1, 3)$ $O=(-1,3)$
$P = (- 1, 4)$ $P=(-1,4)$

$r = \sqrt{{(- 1 + 1)}^{2} + {(4 - 3)}^{2}}$ $r=sqrt((-1+1)^2+(4-3)^2$

$r = \sqrt{0 + 1}$ $r=sqrt(0+1)$

$r = 1 radius of circle$ $r=1 " radius of circle"$

$general equation for circle which it has a center of (a,b) and a radius of r$ $"general equation for circle which it has a center of (a,b) and a radius of r"$
${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $(x-a)^2+(y-b)^2=r^2$

$r = 1 a = - 1 b = 3$ $r=1" "a=-1" "b=3$

${(x + 1)}^{2} + {(y - 3)}^{2} = 1$ $(x+1)^2+(y-3)^2=1$

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How do you write the equation for a circle with center at (-1,3) passes through the point (-1,4)?

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