Circle Theorems - Alternate Segment Theorem Question

Circle Theorem Question Maths GCSE?

Does the alternate segment theorem apply to quadrilaterals, or just triangles? I have a question and it seems like alternate segment theorem would solve everything, but the shape inside the circle is four sided.

How to solve this circle theorem problem?

You are right in your statement.

In addition the two radii from the points of tangency intersects at the center of the circle which formes the CENTRAL angle. Any point along the circle connected to the same points of tangency and still include the arc between them forms an inscribed angle = 1/2 the CENTRAL angle
in the figure the INSCRIBED angle is 39 therefore the Central angle is twice that. Central angle = 78
Now
Another property is the central angle is equal to the deflection angle of the two tangents
Therefore if you extend the one of the tangents past their intersection you form two angle which are supplementary to each other
Supplementary angles = 180
if the deflection of the two tangents = Central angle = 78
Then the INCLUDED angle inside the tangents = 180 - 78
Therefore X = 102

I hope that helped

Circle theorems?

The area of the full circle = pi * r^2
(r being the radius)

A full circle is 360 degrees, therefore 30 degrees represents 1/12 of a full circle.

Area of sector (full slice of pie) = pi * r^2 / 12
(Call it As)

Area of triangle CPR = (1/2) height (perpendicular dropped from C onto chord) * chord
Dropping the perpendicular gives us 2 right-angle triangles dividing triangle CPR in two halves.
Area of triangle (At):
At = (1/2)* r*cos(15) * 2*r*sin(15) = r^2 * sin(15)cos(15)

I use sin(2x) = 2sin(x)cos(x)
backwards, to get
At = (1/2) * r^2 * 2sin(15)cos(15)
At = (1/2) * r^2 * sin(30)
At = (1/2) * r^2 * (1/2) = (1/4) r^2

The area of the slice of pie (As) is the sum of the Area of the triangle (At) and the given area of the "pie crust" (Ac)

As = At + Ac
As - At = Ac

(pi/12) r^2 - (1/4) r^2 = 100
put everything on a common denominator (12)
(pi/12) r^2 - (3/12) r^2 = 1200/12
drop the common denominator (officially, we multiply both sides by 12)
pi * r^2 - 3 * r^2 = 1200
(pi - 3)r^2 = 1200
r^2 = 1200 / (pi - 3)

Once you have the radius, you can find everything else you need.

Circumference = 2 * pi * r
corresponds to 360 degrees.
The 30-degree arc will be 1/12 of that.

Is there an easy way of remembering circle theorems?

Although you're asking about circles in particular, your question is really about learning and understanding the world in general. Basically, everything that you encounter everywhere is arbitrarily complex as you look deeper and deeper. If you accept this, then the only reasonable approach to life, at least math/science/engineering life, is to master general concepts rather than individual mnemonics.

For example, your first rule generalizes to tangent lines being perpendicular to surface normals. This applies to ANY curve or surface, not just circles. So if you remember that one thing then you can scratch that rule off your "special properties of circles" worry list.

The second rule arises from symmetry about the line from the point in question to the circle's center, so having a more general mastery of all the types of symmetries in the world (point symmetry, line symmetry, mirror image, etc) will serve you well. In fact, by symmetry you could generalize the same rule to spheres even though you perhaps haven't officially studied them yet.

An exception to this advice is rule #4 (circle diameter forms hypotenuse) which is a truly lovely property of circles in particular and worth remembering in its own right.

In summary, my advice is to try to remember as few specific things as possible and focus on mastering as many general concepts as possible. Find teachers or tutors or textbooks that present new information in this way, and things will not seem so dauntingly diverse.