Given that x2+y2=16x−26y−133, what is the biggest value that 6x−8y can have?

Note that   x^2+y^2=16x−26y−133  represents a circle.

x^2 - 16x + y^2 + 26y = 133

x^2 - 16x + 64 - 64 + y^2 + 26y + 169 - 169 = 133 (completing the square)

Then (x-8)^2 - 64 + (y+13)^2 - 169 = 133

Simplifying, (x-8)^2 + (y+13)^2 = 133 + 169 + 64  =  366

This circle acts as a constraint on the value of 6x-8y.  Assume that x and y are both on the circle.  Just supposing that x = 10, find y:

(10-8)^2 + (y+13)^2 = 366, or 4 + (y+13)^2 = 366, or (y+13)^2 = 362

This is a quadratic equation that could be solved for y, and the result(s) could be subst. into the expression 6x-8y.

If you were to repeat this exercise several times, for different values of x, you'd come up with various values of 6x-8y and in that way approach (if not find) a definite answer to "Given that x2+y2=16x−26y−133, what is the biggest value that 6x−8y can have?"

Hope someone else can come up with a more elegant approach.

In the first triangle 80+20+p=180 (since sum of all angles of a triangle is 180)so p= 80also in second triangle 55+45+q=180 (since sum of all angles of a triangle is 180)so q=80so we can conclude that p=80 degrees q=80 degrees so p=q=80 degreeshence option b is correct