Find the equation to the circle which has its centre at the point (3,4) and touch the straight line 5x 12y=1

Answer 1

You are missing an operator in your line equation, I will assume you mean
5x + 12y = 1 or 5x + 12y - 1 = 0

we know that the equation must be
(x-3)^2 + (y-4)^2 = r^2 , so the only thing missing is the value of r

using the distance from a point to a line method,
r = |5(3) + 12(4)-1 |/√(3^2 + 4^2) = 62/5

(x-3)^2 + (y-4)^2 = 3844/25