1. **Problem statement:** Given a circle with chord $AB$ of length 16 units, $D$ is the midpoint of chord $AB$, and $C$ is the midpoint of arc $ACB$. The length $CD$ is 4 units. We need to find the radius $r$ of the circle. 2. **Key concepts:** - The midpoint $D$ of chord $AB$ means $AD = DB = 8$ units. - Point $C$ is the midpoint of the arc $ACB$, so $C$ lies on the circle and the arc $ACB$ is divided into two equal arcs by $C$. - The radius $r$ is the distance from the center $O$ of the circle to any point on the circle. 3. **Approach:** - Let $O$ be the center of the circle. - Since $D$ is midpoint of chord $AB$, $OD$ is perpendicular to $AB$. - Let $OD = x$. - Using the right triangle $OAD$, by Pythagoras theorem: $$OA^2 = OD^2 + AD^2$$ $$r^2 = x^2 + 8^2 = x^2 + 64$$ 4. **Using point $C$:** - $C$ lies on the circle, so $OC = r$. - $CD = 4$ units. 5. **Find $OC$ in terms of $x$:** - Since $C$ is midpoint of arc $ACB$, $C$ lies on the circle and the line $OC$ bisects the arc. - The chord $AB$ and point $C$ form an isosceles triangle $OCD$ with $OC = r$ and $OD = x$. - Using the distance $CD = 4$, apply the distance formula in triangle $OCD$: $$CD^2 = OC^2 + OD^2 - 2 \cdot OC \cdot OD \cdot \cos(\theta)$$ But since $OD$ is perpendicular to $AB$, and $C$ lies on the circle, the angle between $OC$ and $OD$ is 90 degrees, so $\cos(90^\circ) = 0$. - Therefore: $$CD^2 = OC^2 + OD^2 = r^2 + x^2$$ Given $CD = 4$, so: $$16 = r^2 + x^2$$ 6. **Combine equations:** - From step 3: $$r^2 = x^2 + 64$$ - From step 5: $$16 = r^2 + x^2$$ Substitute $r^2$ from step 3 into step 5: $$16 = (x^2 + 64) + x^2 = 2x^2 + 64$$ 7. **Solve for $x^2$:** $$2x^2 = 16 - 64 = -48$$ $$x^2 = -24$$ This is impossible for a real radius, so re-examine the assumption about the angle between $OC$ and $OD$. 8. **Correct approach:** - Since $C$ is midpoint of arc $ACB$, $C$ lies on the circle and $OC$ is radius. - $D$ is midpoint of chord $AB$, so $OD$ is perpendicular to $AB$. - Triangle $OCD$ is right angled at $D$ because $OD$ is perpendicular to chord $AB$ and $C$ lies on the circle. - So, $OC^2 = OD^2 + CD^2$. Given $OC = r$, $OD = x$, $CD = 4$: $$r^2 = x^2 + 16$$ From step 3: $$r^2 = x^2 + 64$$ Set equal: $$x^2 + 16 = x^2 + 64$$ This is a contradiction, so the right angle is not at $D$. 9. **Use coordinate geometry:** - Place $D$ at origin $(0,0)$. - Since $D$ is midpoint of $AB$ and $AB=16$, $A=(-8,0)$ and $B=(8,0)$. - Let the center $O$ be at $(0,h)$. - Radius $r = OA = OB = \sqrt{8^2 + h^2} = \sqrt{64 + h^2}$. - Point $C$ lies on the circle and is midpoint of arc $ACB$. - Since $C$ is midpoint of arc $ACB$, $C$ lies on the circle above chord $AB$ at $(0,k)$. - Distance $CD = 4$, so $CD = |k - 0| = 4$, so $k=4$. - Since $C$ lies on circle: $$r^2 = OC^2 = (0 - 0)^2 + (k - h)^2 = (4 - h)^2$$ - Also, from $A$: $$r^2 = 64 + h^2$$ Set equal: $$64 + h^2 = (4 - h)^2 = h^2 - 8h + 16$$ Simplify: $$64 + h^2 = h^2 - 8h + 16$$ $$64 = -8h + 16$$ $$64 - 16 = -8h$$ $$48 = -8h$$ $$h = -6$$ 10. **Calculate radius:** $$r^2 = 64 + h^2 = 64 + (-6)^2 = 64 + 36 = 100$$ $$r = \sqrt{100} = 10$$ **Final answer:** The radius of the circle is $\boxed{10}$ units.