1. **Problem statement:** We have two equal circles with centers $O$ and $O'$ touching at $X$. The line $OO'$ is extended to meet the circle centered at $O'$ at $A$. The line $AC$ is tangent to the circle centered at $O$ at $C$. The segment $O'D$ is perpendicular to $AC$. We need to find the ratio $\frac{DO'}{CO}$. 2. **Key properties and formulas:** - Since the circles are equal, their radii are equal; let the radius be $r$. - The circles touch at $X$, so $OO' = 2r$. - Tangent to a circle is perpendicular to the radius at the point of tangency: $AC \perp OC$. - $O'D$ is perpendicular to $AC$ by given. 3. **Set up coordinate system:** Place $O'$ at the origin $(0,0)$ and $O$ at $(2r,0)$ on the x-axis. 4. **Coordinates of points:** - $A$ lies on the circle centered at $O'$ with radius $r$, so $A$ satisfies $x^2 + y^2 = r^2$. - $A$ lies on the line $OO'$ extended, which is the x-axis, so $A = (-r,0)$ (since $OO' = 2r$, $A$ is on the left side of $O'$). 5. **Find point $C$ on circle centered at $O$:** - Circle $O$: $(x-2r)^2 + y^2 = r^2$. - $AC$ is tangent to circle $O$ at $C$. - Line $AC$ passes through $A(-r,0)$ and $C(x_c,y_c)$. 6. **Slope of $AC$:** Let slope $m = \frac{y_c - 0}{x_c - (-r)} = \frac{y_c}{x_c + r}$. 7. **Tangent condition:** The radius $OC$ is perpendicular to $AC$, so slope of $OC$ is $-1/m$. - Slope of $OC = \frac{y_c - 0}{x_c - 2r} = \frac{y_c}{x_c - 2r}$. - So, $\frac{y_c}{x_c - 2r} = -\frac{1}{m} = -\frac{x_c + r}{y_c}$. 8. **From above:** $$ y_c^2 = -(x_c - 2r)(x_c + r) = -(x_c^2 + r x_c - 2r x_c - 2r^2) = -(x_c^2 - r x_c - 2r^2) $$ 9. **Point $C$ lies on circle $O$:** $$ (x_c - 2r)^2 + y_c^2 = r^2 $$ Substitute $y_c^2$ from step 8: $$ (x_c - 2r)^2 - (x_c^2 - r x_c - 2r^2) = r^2 $$ 10. **Expand and simplify:** $$ (x_c^2 - 4r x_c + 4r^2) - x_c^2 + r x_c + 2r^2 = r^2 $$ $$ -4r x_c + r x_c + 4r^2 + 2r^2 = r^2 $$ $$ -3r x_c + 6r^2 = r^2 $$ 11. **Solve for $x_c$:** $$ -3r x_c = r^2 - 6r^2 = -5r^2 $$ $$ x_c = \frac{5r^2}{3r} = \frac{5r}{3} $$ 12. **Find $y_c^2$ using step 8:** $$ y_c^2 = -(x_c^2 - r x_c - 2r^2) = -\left(\left(\frac{5r}{3}\right)^2 - r \cdot \frac{5r}{3} - 2r^2\right) $$ $$ = -\left(\frac{25r^2}{9} - \frac{5r^2}{3} - 2r^2\right) = -\left(\frac{25r^2}{9} - \frac{15r^2}{9} - \frac{18r^2}{9}\right) = -\left(-\frac{8r^2}{9}\right) = \frac{8r^2}{9} $$ 13. **So, $y_c = \frac{2r \sqrt{2}}{3}$.** 14. **Equation of line $AC$:** Slope $m = \frac{y_c}{x_c + r} = \frac{\frac{2r \sqrt{2}}{3}}{\frac{5r}{3} + r} = \frac{\frac{2r \sqrt{2}}{3}}{\frac{8r}{3}} = \frac{2r \sqrt{2}}{3} \cdot \frac{3}{8r} = \frac{\sqrt{2}}{4}$. 15. **Find point $D$ on $O'A$ such that $O'D \perp AC$:** - $O' = (0,0)$, $A = (-r,0)$, so $O'A$ is on the x-axis. - Since $O'D$ is perpendicular to $AC$ and $O'D$ lies on $O'A$ (x-axis), $D$ has coordinates $(x_D,0)$. 16. **Find foot of perpendicular from $O'$ to $AC$:** Line $AC$ passes through $A(-r,0)$ with slope $m=\frac{\sqrt{2}}{4}$. Equation: $$ y - 0 = \frac{\sqrt{2}}{4}(x + r) \Rightarrow y = \frac{\sqrt{2}}{4}(x + r) $$ 17. **Perpendicular from $O'(0,0)$ to $AC$:** Slope of perpendicular is $-\frac{1}{m} = -\frac{4}{\sqrt{2}} = -2\sqrt{2}$. Equation: $$ y = -2\sqrt{2} x $$ 18. **Find intersection $D$ of these two lines:** $$ \frac{\sqrt{2}}{4}(x + r) = -2\sqrt{2} x $$ Multiply both sides by 4: $$ \sqrt{2}(x + r) = -8\sqrt{2} x $$ Divide both sides by $\sqrt{2}$: $$ x + r = -8 x $$ $$ 9 x = -r \Rightarrow x = -\frac{r}{9} $$ 19. **Coordinates of $D$:** $$ D = \left(-\frac{r}{9}, 0\right) $$ 20. **Calculate lengths:** - $DO' = |x_D - 0| = \frac{r}{9}$. - $CO = \sqrt{(x_c - 2r)^2 + y_c^2}$. 21. **Calculate $CO$:** $$ x_c - 2r = \frac{5r}{3} - 2r = \frac{5r - 6r}{3} = -\frac{r}{3} $$ $$ CO = \sqrt{\left(-\frac{r}{3}\right)^2 + \left(\frac{2r \sqrt{2}}{3}\right)^2} = \sqrt{\frac{r^2}{9} + \frac{8r^2}{9}} = \sqrt{\frac{9r^2}{9}} = r $$ 22. **Ratio:** $$ \frac{DO'}{CO} = \frac{\frac{r}{9}}{r} = \frac{1}{9} $$ **Final answer:** $$ \boxed{\frac{DO'}{CO} = \frac{1}{9}} $$