1. **Problem statement:** We have a square ABCD with side length 9. A circle centered at A with radius AD is drawn, and another circle with diameter DC is drawn. These two circles intersect at point P. We need to find the distance from point P to the line AD, expressed as a fraction $\frac{a}{b}$ in simplest form, and then find $a+b$. 2. **Setup and notation:** - Let the square ABCD have vertices: $A=(0,0)$, $B=(9,0)$, $C=(9,9)$, $D=(0,9)$. - The circle centered at A with radius AD has center $A=(0,0)$ and radius $9$. - The circle with diameter DC has endpoints $D=(0,9)$ and $C=(9,9)$, so its center is the midpoint $M=\left(\frac{0+9}{2}, \frac{9+9}{2}\right)=(4.5,9)$ and radius $\frac{9}{2}=4.5$. 3. **Equations of the circles:** - Circle 1 (center A): $$ (x-0)^2 + (y-0)^2 = 9^2 = 81 $$ - Circle 2 (center M): $$ (x-4.5)^2 + (y-9)^2 = 4.5^2 = 20.25 $$ 4. **Find intersection points P:** Subtract the second circle equation from the first: $$ x^2 + y^2 = 81 $$ $$ (x-4.5)^2 + (y-9)^2 = 20.25 $$ Expanding the second: $$ x^2 - 9x + 20.25 + y^2 - 18y + 81 = 20.25 $$ Simplify: $$ x^2 + y^2 - 9x - 18y + 101.25 = 20.25 $$ $$ x^2 + y^2 - 9x - 18y + 81 = 0 $$ Subtract this from the first circle equation: $$ (x^2 + y^2) - (x^2 + y^2 - 9x - 18y + 81) = 81 - 0 $$ $$ 9x + 18y - 81 = 81 $$ $$ 9x + 18y = 162 $$ Divide by 9: $$ x + 2y = 18 $$ 5. **Substitute $x = 18 - 2y$ into first circle:** $$ (18 - 2y)^2 + y^2 = 81 $$ $$ 324 - 72y + 4y^2 + y^2 = 81 $$ $$ 5y^2 - 72y + 324 = 81 $$ $$ 5y^2 - 72y + 243 = 0 $$ 6. **Solve quadratic:** $$ y = \frac{72 \pm \sqrt{(-72)^2 - 4 \cdot 5 \cdot 243}}{2 \cdot 5} = \frac{72 \pm \sqrt{5184 - 4860}}{10} = \frac{72 \pm \sqrt{324}}{10} = \frac{72 \pm 18}{10} $$ Two solutions: - $$ y_1 = \frac{72 + 18}{10} = \frac{90}{10} = 9 $$ - $$ y_2 = \frac{72 - 18}{10} = \frac{54}{10} = 5.4 $$ 7. **Find corresponding x values:** - For $y=9$: $$ x = 18 - 2(9) = 18 - 18 = 0 $$ - For $y=5.4$: $$ x = 18 - 2(5.4) = 18 - 10.8 = 7.2 $$ 8. **Points of intersection:** - $P_1 = (0,9)$ which is point D (already known) - $P_2 = (7.2, 5.4)$ is the other intersection point P 9. **Distance from P to line AD:** - Line AD is vertical at $x=0$. - Distance from point $(x,y)$ to line $x=0$ is $|x|$. - For $P=(7.2,5.4)$, distance to AD is $7.2$. 10. **Express as fraction:** $$ 7.2 = \frac{72}{10} = \frac{36}{5} $$ - $a=36$, $b=5$, gcd(36,5)=1 11. **Calculate $a+b$:** $$ 36 + 5 = 41 $$ **Final answer:** $\boxed{41}$