Let a circle $$C$$ of radius $$5$$ lie below the $$x$$-axis. The line $$L_1 = 4x + 3y + 2$$ passes through the centre $$P$$ of the circle $$C$$ and intersects the line $$L_2 : 3x - 4y - 11 = 0$$ at $$Q$$. The line $$L_2$$ touches $$C$$ at the point $$Q$$. Then the distance of $$P$$ from the line $$5x - 12y + 51 = 0$$ is ______

A circle $$C$$ of radius 5 lies below the $$x$$-axis. The line $$L_1: 4x + 3y + 2 = 0$$ passes through the centre $$P$$, while $$L_2: 3x - 4y - 11 = 0$$ is tangent to $$C$$ at the point $$Q$$, which is their intersection.

To find $$Q$$, solve $$4x + 3y = -2$$ and $$3x - 4y = 11$$ simultaneously. Multiplying the first equation by 4 and the second by 3 gives $$16x + 12y = -8$$ and $$9x - 12y = 33$$, and adding these yields $$25x = 25$$ so $$x = 1$$ and hence $$y = -2$$. Thus $$Q = (1, -2)$$.

Since $$L_2$$ is tangent to the circle at $$Q$$, the line $$PQ$$ is perpendicular to $$L_2$$. The normal vector to $$L_2: 3x - 4y - 11 = 0$$ is $$(3, -4)$$ with unit vector $$\frac{1}{5}(3, -4)$$. Therefore

$$P = Q \pm 5 \cdot \frac{(3, -4)}{5} = (1, -2) \pm (3, -4),$$

giving $$P = (4, -6)\text{ or }P = (-2, 2).$$

Because the circle lies below the $$x$$-axis, the centre must have $$y < 0$$, so $$P = (4, -6)$$. Verifying on $$L_1$$: $$4(4) + 3(-6) + 2 = 16 - 18 + 2 = 0$$.

Finally, the distance from $$P$$ to the line $$5x - 12y + 51 = 0$$ is

$$d = \frac{\lvert 5(4) - 12(-6) + 51 \rvert}{\sqrt{25 + 144}} = \frac{\lvert 20 + 72 + 51 \rvert}{13} = \frac{143}{13} = 11.$$

The correct answer is 11.