How do you find the center and radius for $x^2+y^2+10x+6y-27=0$ ?

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Answer 1 · 2016-06-22T13:09:08

The center is $(-5,-3)$ and radius is $sqrt61$ .

If the equation is in the form

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

The center is $(h,k)$ and radius is $r$ .

Hence, let us convert the equation $x^2+y^2+10x+6y-27=0$ to this form,

$x^2+10x+y^2+6y=27$

$hArr x^2+10x+25+y^2+6y+9=27+25+9$

$hArr (x+5)^2+(y+3)^2=61$

$hArr (x-(-5))^2+(y-(-3))^2=(sqrt61)^2$

Hence, the center is $(-5,-3)$ and radius is $sqrt61$ .

How do you find the center and radius for x^2+y^2+10x+6y-27=0x^2+y^2+10x+6y-27=0?