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

1 Answer
Sep 21, 2017

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]}