Circle basic laws

Circle basic laws (theorems)

Solved examples

Area of a circle (A)

Circumference of a circle (P)

Diameter is twice the radius: d = 2r

value of π up to 21 decimal places

π = 3.141 592 653 589 793 238 462 ...

Inscribed angles theorem

(1)

All inscribed angles intercepted by the same arc or cord and lies on the same side of the cord are equal.

(2)

Sum of opposite angles drawn from the same cord are. equal to 180°, α + θ = 180°

(3)

If the cord coincides with the diameter of the circle, then the inscribed angle is a right angle and α = θ = 90°.

Central and inscribed angles
If central angle θ and inscribed angle α intercepts the same cord or arc (at the same side) then:

Intersecting secants theorem
If two lines originating from the same point P intersects the circle at two different locations, then:

We can also write:(a + b) a = (c + d) c
If line PD = c is tangent thenc² = (a + b) a

Tangent and secant line theorem
The value of an angle formed by a secant or tangent line drawn from a point P outside the circle equals the half of the difference of the intercepted arcs or central angles.

θ₁ θ₂ are the central angles of the shown arcs.

Subscribed (cyclic) quadrilateral
The sum of the opposite angles in an inscribed quadrilateral are equal to 180°.

α + γ = 180°

β + δ = 180°

Circumcircle in a quadrilateral
The sum of the opposite sides of inscribed (cyclic) quadrilateral are equal.

a + c = b + d

Ptolemy's theorem
The product of the diagonals of a cyclic quadrilateral equals the sum of the products of the opposite sides.

Intersecting chord theorem The product of the two segments created by intersecting of two chords are equal.

Inner angle theorem Inner angle is equal to half the sum of the arcs formed by the intersection of two cords.

Tangent angle theorem
If line L is tangent to the circle, then the angle between the radius connecting circle center and the intersection point with line L forms an angle of 90 degrees (right angle).

Alternate segment angle theorem
If a line L is tangent to the circle, then the angle between the tangent line and the alternates angles are equal.

Two tangents lines to a circle
Two tangents lines that are originating from the same point outside of the circle, are of equal length and the angles α and β are equal.

AB = AC

α = β

Solved examples:

Example 1 - Inscribed angle

AC is the diameter of a circle and the cord BD is perpendicular to the diameter, the length of cord AB is given as 8 cm and the angle ∢BAC = α = 30 degree. Find the length of cord BD and the area of triangle CDE.

We see from the drawing that the inscribed angles α and β have common cord BC therefore they have the same value.

α = β = 30 degree

From the sin equation we can find the length of BE:

BE = AB sinα = 8 * sin30 = 4 cm

Because the cord BD and the diameter AC are perpendicular then BE = ED = 4 cm

And the value of cord BD is: BD = 2 BE = 2 * 4 = 8 cm

Now we can find the length of EC = ED tanβ = 4 tan30 = 2.31 cm

The area of triangle CDE is:

Example 2 - Inscribed angle

Given an inscribed quadrilateral ABCD where side AB is the diameter of the circle. Find angle β as a function of angle α (∢ABD). Find β when α = 27 degree.

Because AD is the diameter then angle ∢ is a right angle and γ = 90 − α.

From the theorem of the sum of the opposite angles of a cyclic quadrilateral is equal to 180 degree we get.

γ + β = 180 degree

90 − α + β = 180 degree

β = 90 + α

β = 90 + 27 = 117 degree

Example 3 - Tangent line

Point O is the center of a circle whose radius is r, from point C outside of the circle two lines were drawn the first is tangent to the circle at point D and the second line cuts the circle at points A and B, such a way that point AD is a dimeter of the circle. another line was drawn from the center to point B, forming an angle α of 65 degrees. Find the length of line AC if the radius equal to 5.

γ is a central angle based on cord AB, the value of angle is:

γ = 180 − α = 180 − 65 = 115 degree.

Now we can find the value of angle β from the isosceles triangle AOB that contains two sides of the radius r (we could use instead the fact that α = 2β complementary angle).

β = (180 - γ) / 2 = (180 − 115) / 2 = 32.5 degree.

Now we can find line AC using cos function.

AC=AD/cos⁡β =2r/cos⁡β =(2∙5)/cos⁡32.5 =11.86

Example 4 - Tangent line

Line AC is tangent to the circle at point A, from point C a straight line was drawn to the center of the circle, the length of line AC is 4 cm and the length of the line from point C to B is 2 cm. Find the lenght of the circle radius.

Because line AC is tangent to the circle then angle OAC is 90 degree, and we can apply the Pythagoras formula:

r^2+(AC) ̅^2=(r+(BC) ̅ )^2

r^2+(AC) ̅^2=r^2+2r(BC) ̅+(BC) ̅^2

2r(BC) ̅=(AC) ̅^2-(BC) ̅^2

r=((AC) ̅^2-(BC) ̅^2)/(2(BC) ̅ )=(144-64)/(2∙8)=5

Example 5 - Inscribed angles

Triangle ABC is inscribed in a circle and O is the center of the circle and the high BD is perpendicular to side AC of the triangle. Prove that angle α is equal to angle β.

figure 1

To prove that angle α = β we shall continue radius BO until it cuts the circle at point E (see figure 2).

Notice that angles γ and δ are inscribed angles with common cord BC therefore they are equal:

γ = δ

Angle BCE is a right angle because it is an inscribed angle with a cord BE coincide with the radius

Figure 2

α = 180 − 90 − γ

β = 180 − 90 − δ

And both angles are the same.

Example 6 - Tangent line

Line BC is tangent to the circle at point C, line BO cuts the circle at point A, given the values of angle α and the radius r. Find the length of line AC and AB.

Because line BC is tangent to the circle then angle BCO is 90 degrees, and we can find the value of angle β

β = 90 − α

Applying the cos law on triangle AOC we have:

(AC) ̅^2=r^2+r^2-2r^r cos⁡β

(AC) ̅^2=r√(2(1-cos⁡β ) )

(AC) ̅=r√(2(1-sin⁡α ) )

The length AB can be solved in different ways, but we will use the intersecting secants theorem to find the length.

(BC) ̅^2=((AB) ̅+r) (AB)

The value of BC is:

Eliminating BC we get:

The solution of this quadratic equation is: