A Short Guide to Circle Geometry

[jamonwindeyer]

Hello all! I've just finished writing a guide to 2 unit geometry. In this guide, I want to add to that by giving a brief run down of what to expect in terms of circle geometry in a HSC/Trial Exam. It's the dark horse of extension exams... It's one of the first things you cover in Year 11 (usually), and lots of students think it won't be covered much in an exam. Wrong . In 2012 it was worth nearly 10% of the marks for the exam. That's the difference between a Band E3 and a Band E4. Read this guide, and it will brush you up on everything you should know. As always, remember to register and ask any questions you have below, and check out the awesome notes available for both 2 Unit and Extension 1.

In Extension One, circle geometry questions are best approached armed with a detailed knowledge of the list of rules (and there is a lot). Look at the question, and ask yourself, what is being shown? If there is tangents, think alternate segments. Are there chords? Think of your chord formulas. Are there distances involved? Think of your ratio formulas. Don't look at the whole thing with every formula in mind; break it down, and consider which formulas COULD be used, and go from there.

Let's introduce the concepts with a few questions.

Example One (HSC 2014) : The points A, B and C lie on a circle with centre O, as shown in the diagram. The size of ∠ACB is 40°. What is the size of ∠AOB? [/b]



This should bring to mind the following ideas:

• Chords which are equidistant from the centre of the circle are the same length and make the same angle with the centre
• The perpendicular from the centre of the circle bisects a chord
• The angle made by a chord at the centre of the circle is twice that made at the circumference (be careful of reflex angles here)

Only that last idea is relevant here, and from this idea, it is easy to see the answer is 80 degrees.

Example Two (HSC 2012): The diagram shows a large semicircle with diameter AB and two smaller semicircles with diameters AC and BC, respectively, where C is a point on the diameter AB. The point M is the centre of the semicircle with diameter AC.
The line perpendicular to AB through C meets the largest semicircle at the point D. The points S and T are the intersections of the lines AD and BD with the smaller semicircles. The point X is the intersection of the lines CD and ST.



a) Explain why CTDS is a rectangle.

Looking at this diagram, the three semicircles should immediately ring alarm bells for the idea that the angle in a semicircle is 90 degrees . I'll save myself the trouble of writing it three times in the proof below, but this fact along with basic straight angle ideas is all you need:

 

Since all angles are 90 degrees, CTDS is a rectangle. I'll skip the rest of that question, it goes more into easy 2U geometry from there.

Example Three (HSC 2014) : In the diagram, AB is a diameter of a circle with centre O. The point C is chosen such that Triangle ABC is acute-angled. The circle intersects AC and BC at P and Q respectively.



a) Why is ∠BAC = ∠CQP?

The answer here is the first of the two key properties of cyclic quadrilaterals, both of which should spring to mind when you spot this diagram.

• The external angle of a cyclic quadrilateral is equal to the opposite internal angle
• The opposite internal angles of a cyclic quadrilateral are supplementary

b) Show that the line OP is a tangent to the circle through P, Q and C.

This one is a little trickier, but as soon as the word tangent is mentioned, you should think angle in alternate segment . BOSTES loves to use it for some reason. The reasoning is as follows.

 Let ∠APO=x degrees
 So ∠OAP=x degrees (since Triangle APO is isosceles, equal radii, and the base angles of an isosceles triangle are equal)
 So, from part(a), ∠CQP=x degrees
 Now if we extend OP beyond the edge of the circle to some point N, the angle ∠CPN=x (vertically opposite at Point P). The solution diagram provided in the marking guidelines shows this nicely.



Now, since ∠CPN (∠CPT in the diagram), is equal to ∠CQP, then the angle in alternate segment theorem is satisfied (the acute angle formed between the tangent and a chord is equal to the acute angle formed by that chord at the circumference). Thus, the line OP is a tangent.

Example Four (HSC 2013): The circles C1 and C2 touch at the point T. The points A and P are on C1. The 3 line AT intersects C2 at B. The point Q on C2 is chosen so that BQ is parallel to PA. Prove that the points Q, T and P are collinear.



This one is again, a little trickier. It doesn't relate to any new ideas (and there are more, which I will summarise in a bit). Instead, it shows that just like the last question, you sometimes have to be proactive, and add in your own information to answer the question. Again, I think the diagram provided by BOSTES does the best job of explaining it, so I'll walk you through it.



We draw a common tangent between the two circles, which remember, allows us to use the angle in alternate segment theorem. Here is the proof.

We let

By angles in alternate segments,

But , alternate angles on PA//BQ

So , which means they are vertically opposite at Point T. As a consequence, Q, T and P must be collinear.

This isn't all of the rules there is. The main ones missed:

• Angles made at the circumference, standing on the same arc are equal. Thus, so are angles made at the centre
• Chords intersect in proportion, meaning that if chords AB and CD intersect at point X:
• A similar formula works for the intersection of a tangent at Point T and an extended chord AB, according to the formula:
• Tangents to a circle from a common, exterior point are equal

These appear less often than the earlier theorems, but they may appear, so be ready for them regardless.

That's it for this guide! Feel free to leave some feedback below, drop a question, give us some tips on how you approach these questions. And another reminder to check out the notes , they are seriously awesome. Have fun!


A GUIDE BY JAMON WINDEYER