If a line has D. Rs. <a+b, b+c, c+a>, then what is the sum of the squares of its D. Cs. ?

X^2 - 4x +(4/2^2) + y^2 + 8y + (8/2^2) = -4 + (4/2^2) + (8/2^2)(x + 2)^2 + (y + 4)^2 = -4 + 4 + 16your center is (-2, -4) and the radius is the sqrt of -4 + 4 + 16 which is 4

So the first thing we do is to multiply the top equation by 3. the reason i want to do this is i notice there will be a 6y on both equations and can possibly add/subtract them. so we get 6y = -3x + 27. adding that and the second equation we get 6x = 27-15 which is 12. divide by 6 and we get 2. ∴ x = 2. substituting 2 into the first equation we get 2y = -2 + 9 which is 7. divide by 2 and we get 7/2. ∴ y = 7/2, x = 2.