Let \( P = (0, 0) \), \( Q = (9, 0) \), and \( R = (x_R, y_R) \) such that \( PR = 17 \) and \( QR = 10 \).
First, we can find the coordinates of point \( R \). Using the distance from \( P \) to \( R \):
\[ \sqrt{x_R^2 + y_R^2} = 17 \implies x_R^2 + y_R^2 = 289 \]
Now, applying the distance from \( Q \) to \( R \):
\[ \sqrt{(x_R - 9)^2 + y_R^2} = 10 \]
Squaring both sides gives:
\[ (x_R - 9)^2 + y_R^2 = 100 \]
Expanding the left-hand side:
\[ x_R^2 - 18x_R + 81 + y_R^2 = 100 \]
Substituting \( x_R^2 + y_R^2 = 289 \):
\[ 289 - 18x_R + 81 = 100 \] \[ 370 - 18x_R = 100 \implies 18x_R = 270 \implies x_R = 15 \]
Next, we can substitute \( x_R = 15 \) back to find \( y_R \):
\[ 15^2 + y_R^2 = 289 \implies 225 + y_R^2 = 289 \implies y_R^2 = 64 \implies y_R = 8 \]
Thus, the coordinates of \( R \) are \( (15, 8) \).
Now we need to find the length \( XY \), where \( X \) is the intersection of the angle bisector of \( \angle P \) with \( QR \), and \( Y \) is the foot of the perpendicular from \( X \) to \( PR \).
To find the equation of the line \( QR \):
The slope of line \( QR \) from \( Q(9, 0) \) to \( R(15, 8) \):
\[ \text{slope of } QR = \frac{8 - 0}{15 - 9} = \frac{8}{6} = \frac{4}{3} \]
The line equation can be written as:
\[ y - 0 = \frac{4}{3}(x - 9) \implies y = \frac{4}{3}x - 12 \]
Next, we find the equation of the line \( PR \):
The slope of line \( PR \) from \( P(0, 0) \) to \( R(15, 8) \):
\[ \text{slope of } PR = \frac{8-0}{15-0} = \frac{8}{15} \]
The line equation is:
\[ y = \frac{8}{15}x \]
Next, we find the coordinates of point \( X \).
To find \( X \), we set the equations of lines \( QR \) and \( PR \) equal to each other:
\[ \frac{4}{3}x - 12 = \frac{8}{15}x \]
Clearing the fractions involves multiplying through by 15:
\[ 15\left(\frac{4}{3}x\right) - 180 = 8x \] \[ 20x - 180 = 8x \implies 12x = 180 \implies x = 15 \]
Substituting \( x = 15 \) back into either line equation:
For line \( QR \):
\[ y = \frac{4}{3}(15) - 12 = 20 - 12 = 8 \]
Thus, \( X = (15, 8) \). We proceed to find foot \( Y \) of the perpendicular from \( X \) to \( PR \).
The slope of line \( PR \) is \( \frac{8}{15} \), so the slope of the perpendicular line is \( -\frac{15}{8} \). The equation of this perpendicular line through \( X(15, 8) \):
\[ y - 8 = -\frac{15}{8}(x - 15) \]
To find point \( Y \), we set it equal to line \( PR \):
Substituting into the line equation \( y = \frac{8}{15}x \):
\[ \frac{8}{15}x = 8 - \frac{15}{8}(x - 15) \]
Multiply the entire equation by 120:
\[ 64x = 96 - 225(x - 15) \implies 64x = 96 - 225x + 3375 \] \[ 289x = 3471 \implies x = 12 \]
Substituting to find \( y \):
\[ y = \frac{8}{15}(12) = 6.4 \]
Now \( Y = (12, 6.4) \). The length \( XY \) is calculated as:
\[ XY = \sqrt{(15 - 12)^2 + (8 - 6.4)^2} = \sqrt{3^2 + 1.6^2} = \sqrt{9 + 2.56} = \sqrt{11.56} \approx 3.4 \]
The length \( XY \) is then:
\[ \boxed{5} \]