In the diagram, the circle has centre $O . O F$ is perpendicular to $D C$ at $F$ and is perpendicular to $A B$ at $E$. If $A B=8, D C=6$ and $E F=1$, determine the radius of the circle.

![](https://cdn.mathpix.com/cropped/2024_04_17_7e8f36a7970c2b706527g-2.jpg?height=401&width=426&top_left_y=1534&top_left_x=1294)
Since $O F$ passes through the centre of the circle and is perpendicular to each of chord $A B$ and chord $D C$, then it bisects each of $A B$ and $D C$. (That is, $A E=E B$ and $D F=F C$.)

To see that $A E=E B$, we could join $O$ to $A$ and $O$ to $B$. Since $O A=O B$ (as they are radii), $O E$ is common to each of $\triangle O A E$ and $\triangle O B E$, and each of these triangles is right-angled, then the triangles are congruent and so $A E=E B$. Using a similar approach shows that $D F=F C$. Since $A E=E B$ and $A B=8$, then $A E=E B=4$.

Since $D F=F C$ and $D C=6$, then $D F=F C=3$.

Join $O$ to $B$ and $O$ to $C$.

Let $r$ be the radius of the circle and let $O E=x$.

Since $\triangle O E B$ is right-angled with $O E=x, E B=4$ and $O B=r$, then $r^{2}=x^{2}+4^{2}$ by the Pythagorean Theorem.

Since $O E=x$ and $E F=1$, then $O F=x+1$.

Since $\triangle O F C$ is right-angled with $O F=x+1, F C=3$ and $O C=r$, then $r^{2}=(x+1)^{2}+3^{2}$ by the Pythagorean Theorem.

![](https://cdn.mathpix.com/cropped/2024_04_17_8aa10a3f608c3bba0bd5g-05.jpg?height=396&width=434&top_left_y=450&top_left_x=1537)

Subtracting the first equation from the second, we obtain $0=\left(x^{2}+2 x+1+9\right)-\left(x^{2}+16\right)$ or $0=2 x-6$ or $x=3$.

Since $x=3$, then $r^{2}=3^{2}+4^{2}=25$ and since $r>0$, we get $r=5$.

ANSWER: 5