Category Archives: Circle Theorems

May 24, 2025 · 10:02 am

Circle Geometry Problem

In the diagram, points $A, B, C, D$ and $Q$ lie on a circle centre $O$ , radius $6$ cm and diameter $BQ, \angle{ABQ}=50^\circ, AB$ is parallel to $DO$ and point $P$ lies on diameter $BQ$ such that $OP=DP=4$ cm.

(a) Find $\angle{BCD}$

(b) Determine the length of $PC$ .

$\angle{BOD}=180^\circ-50^\circ=130^\circ$ (Co-interior angles in parallel lines are supplementary.)

$\angle{BCD}=\frac{1}{2}\times 130=65^\circ$ (Angles subtended by the same arc. The angle at the centre is twice the angle at the circumference.)

$\angle{BCD}=65^\circ.$

Let $PC=x$

From the intersecting chord theorem

$\begin{equation*}4\times x=2\times 10\end{equation}$

$\begin{equation*}4x=20\end{equation}$

$\begin{equation*}x=5\end{equation}$

$PC=5cm$

A chord $AB$ of a circle $O$ is extended to $C$ . The straight line bisecting $\angle{OAB}$ meets the circle at $E$ . Let $\angle{BAE}=x$ . Prove that $EB$ bisects $\angle{OBC}$ .

$\angle{BAO}=2x$ ( $AE$ bisects $\angle {BAO}$ )

$\Delta AOB$ is isosceles ( $AO=B0$ radii of the circle)

$\angle{ABO}=2x$ (Equal angles in isosceles triangle)

Therefore $\angle {AOB}=180^\circ-4x$ (angle sum of a triangle)

$\angle {BEA}=90^\circ-2x$ (angle at the circumference is half angle at the centre)

$\angle{ABE}=180^\circ-x-(90^\circ-2x)=90^\circ+x$ (angle sum of a triangle)

$\angle{CBE}=180^\circ-(90^\circ+x)=90^\circ-x$ (angles on a straight line)

$\angle{OBE}=90^\circ+x-2x=90^\circ-x$

$\angle{OBE}=\angle{CBE}$

Hence, $BE$ bisects $\angle{OBC}$

Intersecting Secant Theorem

$CD$ is a tangent to the circle.

Prove $c^2=a(a+b)$

I am going to add two chords to the circle

$\angle{BDC}=\angle{CAD}$ (angles in alternate segments are congruent)

$\angle{BCD}=\angle{DCA}$ (shared angle)

$\therefore \Delta BDC\cong \Delta{DAC}$ (AA)

Hence

$\frac{DC}{AC}=\frac{BC}{DC}$ (Corresponding sides in similar triangles)

$\frac{c}{a+b}=\frac{a}{c}$

$\therefore c^2=a(a+b)$

1 Comment

Filed under Circle Theorems, Geometry, Year 11 Specialist Mathematics

Tagged as circle geometry, Intersecting Secants Theorem

March 17, 2025 · 12:33 pm

Circle Geometry Question 2

One of my Year 11 Specialist students had this question

Triangle $ABC$ touches the given circle at Points $P, Q, R$ and $S$ only. The secant $BW$ touches the circle at $V$ and $W$ .

Diagram not drawn to scale

(a) Determine the lengths of the line segments marked $x, y$ and $z$ , leaving your answers as exact values.

(b) If the length of the line segment $QW$ is 4 units, determine the exact radius of the circle.

(a) We are going to use the Intersecting Secant Theorem – the tangent version

Hence, we have

$\begin{equation*}30^2=25(25+x+6)\end{equation}$

$\begin{equation*}900=25(31+x)\end{equation}$

$\begin{equation*}x=5\end{equation}$

Then we can use the intersecting chord theorem to find $y$ .

$\begin{equation*}10\times y=6 \times x\end{equation}$

$\begin{equation*}10y=30\end{equation}$

$\begin{equation*}y=3\end{equation}$

Back to the Intersecting Secant Theorem to find $z$

$\begin{equation*}z^2=4\times 17\end{equation}$

$\begin{equation*}z=2\sqrt{17}\end{equation}$

(b)

$QW$ is part of a 3-4-5 triangle, therefore $\angle{Q}=90^\circ$

This is definitely the case of the diagram not being drawn to scale. If $\angle{Q}=90^\circ$ , then the purple line must be the diameter.

We can use pythagoras to find the length of the diameter

$\begin{equation*}(2r)^2=13^2+4^2\end{equation}$

$\begin{equation*}4r^2=185\end{equation}$

$\begin{equation*}r=\frac{\sqrt{285}}{2}\end{equation}$

The radius of the circle is $\frac{\sqrt{285}}{2}$

Circle Geometry Question

In the above diagram $O$ is the centre of the larger circle. $A, B,D$ and $E$ are points on the circumference of the larger circle. $A, C, E$ and $0$ are points on the circumference of the smaller circle. Show that $\angle{CAB}=\angle{ABC}$ . $AB, AC$ and $BC$ are straight lines.