What are the center and radius of the circle defined by the equation #x^2+y^2-6x+8y+21=0#?

Question

9419 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-23T05:55:19

The center is #(3,-4)# and the radius is #=2#

We need #(a+b)^2=a^2+2ab+b^2#

Let's rearrange the equation as

#x^2-6x+y^2+8y=-21#

We can now complete the squares

#x^2-6x+9+y^2+8y+16=-21+9+16#

We can now factorise

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

This is the equaton of a circle, center #(3,-4)# and radius #=2#

graph{(x-3)^2+(y+4)^2-4=0 [-6.42, 9.38, -7.07, 0.83]}

Answer 2 · 2016-12-23T06:00:33

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

#x^2+y^2-6x+8y+21=0# can be written as

#x^2-6x+y^2+8y=-21#

or #x^2-2xx3xx x+3^2+y^2+2xx4xxy+4^2=-21+3^2+4^2#

or #(x-3)^2+(y+4)^2=-21+9+16#

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

This is the locus of a point which moves so that its distance from point #(3,-4)# is always #2#.

Hence this is the equation of a circle whose center is #(3,-4)# and radius is #2#.
graph{x^2+y^2-6x+8y+21=0 [-2.043, 7.957, -6.12, -1.12]}