How do you find the equation of a circle in standard form given $x^2+y^2-10x+6y+30=0$ ?

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Answer 1 · 2017-01-23T09:38:58

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

The standard form of equation of a circle is the center-radius form of its equation and it is in the format

$(x-h)^2+(y-k)^2=r^2$ , whose center is $(h,k)$ and radius is $r$ .

To convert $x^2+y^2-10x+6y+30=0$ into standard form, we should group $x$ -terms and $y$ -terms separately as follows:

$x^2-10x+y^2+6y=-30$ and now competing squares, this becomes

$(x^2-2xx5xx x+5^2)+(y^2+2xx3xxy+3^2)=-30+5^2+3^2$

or $(x-5)^2+(y+3)^2=-30+25+9$

i.e. $(x-5)^2+(y-(-3))^2=4$

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

Hence, center is $(5,-3)$ and radius is $2$
graph{x^2+y^2-10x+6y+30=0 [-5, 15, -7.96, 2.04]}

How do you find the equation of a circle in standard form given x^2+y^2-10x+6y+30=0x^2+y^2-10x+6y+30=0?