How do you find the center and radius for #x^2 - 8x + y^2 - 4y – 5 = 0#?

Question

5950 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-10T09:22:21

Center is #(4,2)# and radius is #5#

In the equation of a circle in form

#x^2+y^2+2gx+2fy+c=0#

The center of circle is at #(-g,-f)# and radius is #sqrt(g^2+f^2-c)#

In equation #x^2-8x+y^2-4y-5=0#, #g=-4#, #f=-2# and #c=-5#

Hence, center is #(4,2)# and radius is

#sqrt(4^2+2^2-(-5))=sqrt25=5#

graph{x^2-8x+y^2-4y-5=0 [-7.08, 15.42, -3.345, 7.905]}