How do you write an equation for a circle with center(4,7) and passing through (3,-2)?

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Answer 1 · 2016-06-29T15:27:36

${(x - 4)}^{2} + {(y - 7)}^{2} = 82$ $(x-4)^2+(y-7)^2=82$

Using the center - radius form

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $(x-h)^2+(y-k)^2=r^2$

The radius $r$ $r$ may be computed using the center and the given point

$r = \sqrt{{(7 - - 2)}^{2} + {(4 - 3)}^{2}}$ $r=sqrt((7--2)^2+(4-3)^2)$

$r = \sqrt{82}$ $r=sqrt82$

Let us formulate the equation

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $(x-h)^2+(y-k)^2=r^2$

${(x - 4)}^{2} + {(y - 7)}^{2} = {(\sqrt{82})}^{2}$ $(x-4)^2+(y-7)^2=(sqrt(82))^2$

${(x - 4)}^{2} + {(y - 7)}^{2} = 82$ $(x-4)^2+(y-7)^2=82$

God bless....I hope the explanation is useful.