What is the center and radius of #x^2+4x+y^2-6y-3=0#?

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Answer 1 · 2018-06-01T12:26:57

Complete the square

#(x+2)^2-4+(y-3)^2-9=3#

Get it into the general equation for a circle
#(x-a)^2+(y-b)^2=r^2# gives centre (a,b) and radius r

#(x+2)^2+(y-3)^2=16#

Centre (-2,3) radius 4

Answer 2 · 2018-06-01T12:33:07

Center is at #(-2,3)# and radius is #4# unit.

# x^2+4 x+ y^2 -6 y -3=0 # or

# x^2+4 x +4 + y^2 -6 y+9 =4+9+3 # or

# (x+2)^2+ (y-3)^2 =4^2 # The center-radius form of the circle

equation is #(x – h)^2 + (y – k)^2 = r^2#, with the center being at

the point #(h, k)or (-2,3)# and the radius being #r =4#.

So, center is at #(-2,3)# and radius is #4# unit

graph{x^2+4x+y^2-6 y-3=0 [-20, 20, -10, 10]}