Question

What is the center, radius, and intercepts of the circle with the given equation `x^2 + y^2 +2x - 6y +1 =0`?

Answer 1

centre `(-1,3)`

radius`r=sqrt9=3`

intercepts: `(-1,0), (0,3+-2sqrt2)`

Explanation:

we first complete the square

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

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

`(x^2+2x+1^2)+(y^2-6y+(-3)^2)+1-1^2-(-3)^2=0`

`(x+1)^2+(y-3)^2=9`

compare to standard eqn circle centre`(a,b), "radius "r`

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

centre `(-1,3)`

radius`r=sqrt9=3`

intercepts

on x-axis `y=0`

`(x+1)^2+(-3)^2=9`

(x+1)^2=9-(-3)6^2=0

=>x+1=0

`x=-1`

only one point on the x-axis

on y-axis `x=0`

`(0+1)^2+(y-3)^2=9`

`(y-3)^2=8`

`y-3=+-sqrt8=+-2sqrt2`

`y=3+-2sqrt2`

the graph below shows the circle

graph{x^2+y^2+2x-6y+1=0 [-10, 10, -5, 5]}