A circle is tangent to three sides of a rectangle having side lengths 2 and 4 as shown. A diagonal of the rectangle intersects the circle at points $A$ and $B$. The length of $A B$ is
(A) $\sqrt{5}$
(B) $\frac{4 \sqrt{5}}{5}$
(C) $\sqrt{5}-\frac{1}{5}$
(D) $\sqrt{5}-\frac{1}{6}$
(E) $\frac{5 \sqrt{5}}{6}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-140.jpg?height=217&width=379&top_left_y=1512&top_left_x=1320)
A circle is tangent to three sides of a rectangle having side lengths 2 and 4 as shown. A diagonal of the rectangle intersects the circle at points $A$ and $B$. The length of $A B$ is
(A) $\sqrt{5}$
(B) $\frac{4 \sqrt{5}}{5}$
(C) $\sqrt{5}-\frac{1}{5}$
(D) $\sqrt{5}-\frac{1}{6}$
(E) $\frac{5 \sqrt{5}}{6}$

## Solution 1

Of the many ways to solve this problem perhaps the easiest is to use a Cartesian co-ordinate system. There are two choices for the origin, the centre of the circle and the bottom left vertex of the rectangle. The first solution presented uses the centre of the circle as the origin where the axes are lines drawn parallel to the sides of the rectangle through $(0,0)$. The equation of the circle is now $x^{2}+y^{2}=1$ and the equation of the line containing the diagonal is $y+1=\frac{1}{2}(x+1)$ or $x-2 y=1$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-283.jpg?height=352&width=567&top_left_y=1423&top_left_x=1256)

Equation of Line: $y+1=\frac{1}{2}(x+1) \Leftrightarrow x-2 y=1$ Equation of Circle: $x^{2}+y^{2}=1$

Solving to find the intersection points of the line and circle gives, $(2 y+1)^{2}+y^{2}=1$

$$
\begin{aligned}
& 5 y^{2}+4 y=0 \\
& y(5 y+4)=0 \\
& y=0 \text { or } y=-\frac{4}{5}
\end{aligned}
$$

If we substitute these values of $y$ into $x-2 y=1$ we find, The points of intersection are $A\left(\frac{-3}{5},-\frac{4}{5}\right)$ and $B(1,0)$.

The length of $A B$ is $\sqrt{\left(1+\frac{3}{5}\right)^{2}+\left(\frac{4}{5}\right)^{2}}=\sqrt{\frac{80}{25}}=\frac{4 \sqrt{5}}{5}$.

Note: The other suggested choice for the origin would involve using the equation of the line $y=\frac{1}{2} x$ and the circle $(x-1)^{2}+(y-1)^{2}=1$.

## Solution 2

The diameter, $B G$, is parallel to the sides of the rectangle and has its endpoint at $B$ which is the centre of the rectangle.

Draw $O C \perp A B$ where $O$ is the centre of the circle and by chord properties of a circle, $C B=\frac{1}{2} A B$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-284.jpg?height=298&width=545&top_left_y=610&top_left_x=1278)

Since $O B \| D E, \angle B D E=\angle C B O=\alpha$.

Thus, $\triangle O B C$ is similar to $\triangle F D E$.

Since $D F=2 \sqrt{5}, \frac{1}{2 \sqrt{5}}=\frac{C B}{4}$ and $C B=\frac{2}{5} \sqrt{5}$.

The length of $A B$ is $2\left(\frac{2}{5} \sqrt{5}\right)$ or $\frac{4}{5} \sqrt{5}$.

ANSWER: (B)