What is the equation of a circle with center (-3, -5) and radius 4?

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What is the equation of a circle with center (-3, -5) and radius 4?

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Answer 1 · 2016-03-07T18:13:27

${(x + 3)}^{2} + {(y + 5)}^{2} = 16$ $(x+3)^2+(y+5)^2=16$

Explanation:

A general circle with centre $(a, b)$ $(a,b)$ and radius $r$ $r$ has equation ${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $(x-a)^2+(y-b)^2=r^2$ .

So in this particular case the equation is

${(x + 3)}^{2} + {(y + 5)}^{2} = 16$ $(x+3)^2+(y+5)^2=16$

Answer 2 · 2016-03-07T18:18:00

$x^{2} + y^{2} + 6 x + 10 y + 9 = 0$ $x^2+y^2+6x+10y+9=0$

Explanation:

${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $(x-a)^2+(y-b)^2=r^2$
$a = - 3 b = - 5 r = 4$ $a=-3" "b=-5" "r=4$
${(x + 3)}^{2} + {(y + 5)}^{2} = 4^{2}$ $(x+3)^2+(y+5)^2=4^2$
${(x + 3)}^{2} + {(y + 5)}^{2} = 16$ $(x+3)^2+(y+5)^2=16$
$x^{2} + 6 x + 9 + y^{2} + 10 y + 25 = 16$ $x^2+6x+9+y^2+10y+25=16$
$x^{2} + y^{2} + 6 x + 10 y + 9 = 0$ $x^2+y^2+6x+10y+9=0$

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What is the equation of a circle with center (-3, -5) and radius 4?

2 Answers

Explanation:

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