SOLUTION: the question asks you to find all points (x,y) with x=y that are 4 units from (1,3). this is a circle problem, and i have set the equation up like this: (x-1)2 + (y-3)2 = 16

Algebra ->  Coordinate-system -> SOLUTION: the question asks you to find all points (x,y) with x=y that are 4 units from (1,3). this is a circle problem, and i have set the equation up like this: (x-1)2 + (y-3)2 = 16       Log On


   

Question 158975This question is from textbook college algebra and trig
: the question asks you to find all points (x,y) with x=y that are 4 units from (1,3).
this is a circle problem, and i have set the equation up like this:
(x-1)2 + (y-3)2 = 16 (where the "2" s are squares, obviously)
from here, i have tried several ways to solve the equation, but have run into snags. Thanks This question is from textbook college algebra and trig

Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
i have no idea if this is correct but i did find a solution.
here's how............
%28x-1%29%5E2+%2B+%28y-3%29%5E2+=+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 %28x-1%29%5E2+%2B+%28x-3%29%5E2+=+16.
that became 2%2Ax%5E2+-+8%2Ax+%2B+10+=+16 which became x%5E2-4%2Ax%2B5=8 by dividing both sides of the equation by 2.
that became x%5E2-4%2Ax-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=%28%28-b%29%2Bsqrt%28b%5E2-4%2Aa%2Ac%29%29%2F%282%2Aa%29 and x=%28%28-b%29-sqrt%28b%5E2-4%2Aa%2Ac%29%29%2F%282%2Aa%29.
this provided values for x of +4.64575.... and -.64575...
plugging these values into the original equation, i got %284.64575-1%29%5E2+%2B+%284.64575-3%29%5E2+=+16, for the larger x value, and i got %28-.64575-1%29%5E2+%2B+%28-.64575-3%29%5E2+=+16 for the smaller x value.
both these substitutions into the equation %28x-1%29%5E2+%2B+%28x-3%29%5E2+=+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.