Problem of the Week
Problem E and Solution
One Circle

Problem

A circle with centre $C$ has diameter $AE$. A line segment is drawn from a point $B$ on the circumference of the circle to $D$ on $CE$ such that $BD$ is perpendicular to $CE.$

If $AB=24$ and $CD=2$, determine the radius of the circle.

Solution

Solution 1

Since $C$ is the centre of the circle that passes through $A$ and $B$, then $AC$ and $BC$ are radii. Let $AC=BC=x$ and $BD=y.$

Since $BD\perp CE$, we have $BD\perp AE$. It follows that $\mathrm{\angle }ADB=90\mathrm{°}.$ Using the Pythagorean Theorem in $\mathrm{△}ADB$, $$\begin{array}{rl}(x+2{)}^2+y^2& ={24}^2\\ \text{(1)}& y^2& =576-(x+2{)}^2\end{array}$$

Using the Pythagorean Theorem in $\mathrm{△}CDB$, $$\begin{array}{rl}{2}^2+y^2& =x^2\\ \text{(2)}& y^2& =x^2-4\end{array}$$

From equations $(1)$ and $(2)$ we can conclude, $$\begin{array}{rl}576-(x+2{)}^2& =x^2-4\\ 576-x^2-4x-4& =x^2-4\\ 0& =2x^2+4x-576\\ 0& =x^2+2x-288\\ 0& =(x+18)(x-16)\end{array}$$ Thus, $x=16$, since $x>0$. Therefore, the radius of the circle is $16.$

Solution 2

Since $C$ is the centre of the circle that passes through $A$ and $B$, then $AC$ and $BC$ are radii. Let $AC=BC=x.$ Let $F$ be the point on $AB$ such that $CF$ is perpendicular to $AB$. Since $\mathrm{△}ABC$ is isosceles, $F$ is also the midpoint of $AB.$

Since $\mathrm{\angle }BAC=\mathrm{\angle }BAD$ and $\mathrm{\angle }AFC=\mathrm{\angle }ADB=90\mathrm{°}$, it follows that $\mathrm{△}AFC\sim \mathrm{△}ADB$. Therefore, $$\begin{array}{rl}\frac{AC}{AF}& =\frac{AB}{AD}\\ \frac{x}{12}& =\frac{24}{x+2}\\ x(x+2)& =288\\ x^2+2x-288& =0\\ (x+18)(x-16)& =0\end{array}$$

Thus, $x=16$, since $x>0.$ Therefore, the radius of the circle is $16.$