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

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

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

So in this particular case the equation is

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

Answer 2 · 2016-03-07T18:18:00

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

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