Lines Tangent to a Circle

Since the line joining the two points of tangency is perpendicular to the given two lines it will have slope $\dfrac{5}{2}$ . Since it passes through the origin, the slope of this line will also equal $\dfrac{y}{x}$ , and so

$\dfrac{y}{x} = \dfrac{5}{2} \Longrightarrow y = \dfrac{5}{2}x$ .

Since $(x,y)$ lies on the unit circle, we also know that $x^{2} + y^{2} = 1$ . So upon substitution we have that

$x^{2} + \dfrac{25}{4}x^{2} = 1 \Longrightarrow x^{2} = \dfrac{4}{29} \Longrightarrow x = \dfrac{2}{29}\sqrt{29}$ .

This implies that $y = \dfrac{5}{29}\sqrt{29}$ , and so $x + y = \dfrac{7}{29}\sqrt{29}$ .

Thus $a = 7, b = 29, c = 29$ and $a + b + c = \boxed{65}$ .

This problem can be solved by taking the derivative of the positive half of the unit circle ( $- \dfrac{x}{\sqrt{1-x^2}}$ ), setting this equal to $- \dfrac{2}{5}$ , and solving for $x$ to obtain $x=\dfrac{2 \sqrt{29}}{29}, y=\dfrac{5 \sqrt{29}}{29}$ and $x+y=\dfrac{7 \sqrt{29}}{29}$ . However, this method is very ugly and long, so I would like to know if there are any better solutions (or anything that doesn't involve calculus).