What are the center and radius of the circle defined by the equation #x^2+y^2-6x+10y+25=0#?

Question

8877 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-01T02:41:28

The given equation represents a circle with center at #(3,-5)# and radius #3#.

#x^2+y^2-6x+10y+25=0# can be written as

#x^2-6x+y^2+10y+25=0# or

#x^2-6x+9+y^2+10y+25=9# or

#(x-3)^2+(y+5)^2=3^2# or

#(x-3)^2+(y-(-5))^2=3^2#

This is the locus of a point which moves so that it moves at constant distance of #3# from the point #(3,-5)#.

Hence the given equation represents a circle with center at #(3,-5)# and radius #3#.
graph{x^2+y^2-6x+10y+25=0 [-7.92, 12.08, -9.52, 0.48]}