Question 158975
i have no idea if this is correct but i did find a solution.
here's how............
{{{(x-1)^2 + (y-3)^2 = 16}}} is a good starting point since that is the equation for the distance between the point (1,3) and any other point that is 4 units away from it.
this is also the equation for a circle with (1,3) at the center and a radius of 4 since all x and y values will be on that circle.
the problem stated that x = y so i substituted x for y in the equation to get {{{(x-1)^2 + (x-3)^2 = 16}}}.
that became {{{2*x^2 - 8*x + 10 = 16}}} which became {{{x^2-4*x+5=8}}} by dividing both sides of the equation by 2.
that became {{{x^2-4*x-3=0}}} by subtracting 8 from both sides of the equation.
i couldn't find any factors that would make it easy so i solved by using the quadratic formula of {{{x=((-b)+sqrt(b^2-4*a*c))/(2*a)}}} and {{{x=((-b)-sqrt(b^2-4*a*c))/(2*a)}}}.
this provided values for x of +4.64575.... and -.64575...
plugging these values into the original equation, i got {{{(4.64575-1)^2 + (4.64575-3)^2 = 16}}}, for the larger x value, and i got {{{(-.64575-1)^2 + (-.64575-3)^2 = 16}}} for the smaller x value.
both these substitutions into the equation {{{(x-1)^2 + (x-3)^2 = 16}}} yielded the identity 16 = 16 proving these were correct values for x and, since y = x, for y as well.
at least that's what i think.
i don't know if there's other values to satisfy the equation as well.  i think i got at least 2.