Review question

# Can we find the areas enclosed between a circle and its tangent? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8946

## Suggestion

In the diagram below is sketched the circle with centre $(1,1)$ and radius $1$ and a line $L$. The line $L$ is tangential to the circle at $Q$. Further, $L$ meets the $y$-axis at $R$, and the $x$-axis at $P$ in such a way that the angle $OPQ$ equals $\theta$ where $0<\theta<\dfrac{\pi}{2}$.

1. Show that the co-ordinates of $Q$ are $(1+ \sin \theta, 1 + \cos \theta),$ and that the gradient of $PQR$ is $- \tan \theta$.

Could we add the radius of the circle which meets the tangent to the diagram? What angles do we know now?

1. The region bounded by the circle, the $x$-axis and $PQ$ has area $A(\theta)$; the region bounded by the circle, the $y$-axis and $QR$ has area $B(\theta)$. (See diagram.)

Explain why $A(\theta) = B\left(\dfrac{\pi}{2} - \theta\right)$ for any $\theta$.

If we draw two diagrams, one with angle $\theta$ at $P$, and the other with angle $\left(\dfrac{\pi}{2} - \theta\right)$ at $P$, then what do we notice?

Calculate $A\left(\dfrac{\pi}{4}\right)$.

Could we sketch a new diagram with $\theta=\dfrac{\pi}{4}$? Can we calculate the length $OP$?

1. Show that $A \left(\frac{\pi}{3} \right) = \sqrt{3} - \frac{\pi}{3}.$

Could we sketch a new diagram with $\theta=\dfrac{\pi}{3}$?

Maybe $A \left(\dfrac{\pi}{3} \right)$ could be found by finding the area of two right-angled triangles and subtracting part of the area of the circle?