Question

How can I solve it? Coordinate geometry. I need the circle's equation from this:

`x^2 + y^2 - 2x + 6y +1 = 0`

Answer 1

The circle's equation is `(x-x_c)^2+(y-y_c)^2=r^2` where `(x_c,y_c)` is the circle's center

Explanation:

Remember: `(x+a)^2=x^2+2*x*a+a^2`
`x^2-2*x+y^2-6*y+1=0`
1. Focus on x:
`(x-2)^2=x^2-2*x+1 rArr x^2-2*x=(x-2)^2-1`
insted of `x^2-2*x` write `(x-2)^2-1`
`(x-2)^2-1 +y^2-6*y+1=0`
2. Do the same for y:
`(y-3)^2=y^2-2*3*y+9=y^2-6*y+9 rArr y^2-6*y=(y-3)^2-9`
Put it in your equation:
`(x-2)^2-1 +(y-3)^2-9+1=0`
`(x-2)^2+(y-3)^2-9=0`
`(x-2)^2+(y-3)^2=9`
This is the circle's equation:
`(x-2)^2+(y-3)^2=9`
Radius:`r=3`
Center: `(2,3)`