Question 1146433
it has to be the equation of a circle.
For a circle of radius K=any positive number,
centered at (3,4), the equation would be
{{{(x-3)^2+(y-4)^2=K^2}}} , which "simplifies" to
{{{x^3+y^2-6x-8y=K^2-3^2-4^2}}}
The answer is obviously
​c. x2 + y2 - 6x - 8y = 0,
unless they were tricking you,
and there is no positive number K such that {{{K^2-3^2-4^2}}} is {{{0}}} .
({{{5^2-3^2-4^2}}}  is zero).