How to do: Find the centre and radius of the circle with equation: x^2+y^2+4x+6y+12=0

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A circle with center coordinates (a,b) and radius r can be expressed as;

`(x-a)^2+(y-b)^2 = r^2`

`x^2-2ax+a^2+y^2-2by+b^2 = r^2`

`x^2+y^2-2ax-2by+a^2+b^2 = r^2`

`x^2+y^2-2ax-2by+(a^2+b^2-r^2) = 0` ---(1)

 

Now let us arrange `x^2+y^2+4x+6y+12=0` same as (1)

`x^2+y^2+4x+6y+12=0 `

`x^2+y^2-2(-2)x-2(-3)y+12 = 0`

Now this is identical with (1)

a = -2

b = -3

`(a^2+b^2-r^2) = 12`

`(-2)^2+(-3)^2-r^2 = 12`

           ` 4+9-r^2 = 12`

                    `r^2 = 13-12`

                      ` r = 1` since r>0

 

So the center is (-2,-3) and radius is 1.

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