Question

For what set of values of K in the equation x^2 + y^2 - 2x - 4y = k, will the equation represent a real circle?

Answer 1

circle's equation :
(x-xcenter)^2 + (y-ycenter)^2 = r^2

so now we will complete the square in your equation

x^2 -2x +y^2 -4y = k

for x^2-2x to complete the square we need to add 1 and so we have to subtract it later so we wont change anything :

x^2-2x+1-1+y^2-4y = k
(x-1)^2 +y^2-4y = k+1

for y^2-4y we have to add 4 and the subtract 4

(x-1)^2 +y^2-4y+4-4 = k+1
(x-1)^2 + (y-2)^2 = k + 5

so k+5 > 0
so k > -5