A generalization of Butterfly Theorem?

See Haruki's lemma here.

Assuming the question is correct and that the answer is the same for any $MR$ and $NQ$ through point $P$ , pick $M$ and $Q$ so that they are both on $A$ and pick $R$ and $N$ so that they are both on $B$ . Then $X$ is also on $A$ and $Y$ is also on $B$ so that $XP = AP = 8$ and $YP = PB = 15$ .

Then $\frac{1}{XP} - \frac{1}{YP} = \frac{1}{8} - \frac{1}{15} = \frac{7}{120}$ , so $a = 7$ and $b = 120$ and $a + b = \boxed{127}$ .

---

To prove it without the assumption, place the diagram on a coordinate graph such that $P$ is at the origin and $A$ and $B$ are on the $x$ -axis. Let $a = AP$ and $b = PB$ , let $c$ be the shortest distance between the center of the circle to segment $AB$ , let $MR$ have a slope of $m_1$ so that its equation is $y = m_1x$ and let $QN$ have a slope of $m_2$ so that its equation is $y = m_2x$ .

Then the circle has a center of $(\frac{b-a}{2}, c)$ and its radius $r$ is the distance between this point and $(b, 0)$ , so that $r^2 = c^2 + \frac{(a + b)^2}{4}$ . The equation of the circle is then $(x - \frac{b-a}{2})^2 + (y - c)^2 = c^2 + \frac{(a + b)^2}{4}$ .

Combining the circle equation and $y = m_1x$ gives an $x$ -coordinate for $R$ of $R_x = \frac{b - a + 2m_1c + \sqrt{(b - a + 2m_1c)^2 - 4ab(m_1^2 + 1)}}{2(m_1^2 + 1)}$ and a $x$ -coordinate for $M$ of $M_x = \frac{b - a + 2m_1c - \sqrt{(b - a + 2m_1c)^2 - 4ab(m_1^2 + 1)}}{2(m_1^2 + 1)}$ . Combining the circle equation and $y = m_2x$ gives an $x$ -coordinate for $N$ of $N_x = \frac{b - a + 2m_2c + \sqrt{(b - a + 2m_2c)^2 - 4ab(m_2^2 + 1)}}{2(m_2^2 + 1)}$ and a $x$ -coordinate for $Q$ of $Q_x = \frac{b - a + 2m_2c - \sqrt{(b - a + 2m_2c)^2 - 4ab(m_2^2 + 1)}}{2(m_2^2 + 1)}$ . Note that $M_x + R_x = \frac{b - a + 2m_1c}{m_1^2 + 1}$ , $Q_x + N_x = \frac{b - a + 2m_2c}{m_2^2 + 1}$ , $M_xR_x = \frac{-ab}{m_1^2 + 1}$ , and $Q_xN_x = \frac{-ab}{m_2^2 + 1}$ .

The equation of the line through $RN$ is $y - m_1R_x = \frac{m_1R_x - m_2N_x}{R_x - N_x}$ , and when $y = 0$ , $x = \frac{R_xN_x(m_1 - m_2)}{m_1R_x - m_2N_x} = YP$ . Similarly, the equation of the line through $QM$ is $y - m_2Q_x = \frac{m_2Q_x - m_1M_x}{Q_x - M_x}$ , and when $y = 0$ , $x = \frac{Q_xM_x(m_2 - m_1)}{m_2Q_x - m_1M_x} = -XP$ .

Therefore, $\frac{1}{XP} - \frac{1}{YP} = \frac{m_2Q_x - m_1M_x}{Q_xM_x(m_1 - m_2)} - \frac{m_1R_x - m_2N_x}{R_xN_x(m_1 - m_2)}$ $=$ $\frac{N_xQ_xR_xm_2 - M_xN_xR_xm_1 - M_xQ_xR_xm_1 + M_xN_xQ_xm_2}{M_xN_xQ_xR_x(m_1 - m_2)}$ $=$ $\frac{1}{m_1 - m_2}\bigg(\big(\frac{M_x + R_x}{M_xR_x}\big)m_2 - \big(\frac{Q_x + N_x}{Q_xN_x}\big)m_1\bigg)$ , and using the relationships for $M_x$ , $N_x$ , $Q_x$ , and $R_x$ above, this simplifies to $\frac{1}{XP} - \frac{1}{YP} = \frac{1}{a} - \frac{1}{b}$ .

In this problem, $a = 8$ and $b = 15$ , so $\frac{1}{XP} - \frac{1}{YP} = \frac{1}{8} - \frac{1}{15} = \frac{7}{120}$ . Therefore, $a = 7$ , $b = 120$ , and $a + b = \boxed{127}$ .