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Circle

The circle is the set of all points of the plane that are equally distanced from one fixed point of the same plane that we call center of the circle.

The distance from the center to any point on the circle is called the radius of the circle.

radius

Chord is the line that connects two different points that lies on it.

chord

The longest chord is called the diameter. Center lies on the diameter. The length of a diameter is equal to 2 * r.

 

diameter

Sector is an enclosed part of a circle which contains two radius and a part of a circumference.

sector

Segment is a part of a circle enclosed by a chord and a part of the circumference.

segment

Secant is the line that intersects the circle in two points and tangent is a line that has exactly one point in common with the circle.

secanttangent

 

Concentric circles have the same center but different radius.

concentric circles

If r is the length of the circle, then its circumference is given with the formula:

C = 2 * r * \pi,

and area with

A = r^2 * \pi

From this we can see that the number \pi is defined as the ration of the length of circumference and its diameter.

 

Length of the arc and area of a sector

Firstly, let’s observe circle on which lie two different points A and B. Those points divide it in two different arcs. Their union is the whole circle.

two different points on the circle

For denoting arcs we have to refer to the positive direction – opposite of the direction of the clocks hands.

This means that \hat{AB} \not= \hat{AB}

We can draw angle whose axis is the center of the circssle and whose legs go through the points A and B. this angle is called the central circle.

central circleWhole circle has and the circumference is 360^{\circ}. This means that we can make the following ratio:

l ( 1^{\circ}) = \frac{2*r*\pi}{360^{\circ}}

i.e.

l ( 1^{\circ}) = \frac{r*\pi}{180^{\circ}}

If the central angle has \alpha degrees; than the length of the arc is α times greater than the arc that matches the 1^{\circ} angle.

l (\alpha) = \frac{r*\pi}{180^{\circ}} * \alpha

In the similar way we can find the formula for the area of the sector.

P (1^{\circ}) = \frac{r^2 * \pi}{360^{\circ}}

P (\alpha) = \frac{r^2 * \pi}{360^{\circ}} * \alpha

 

Central and inscribed angles

Every angle whose vertex is a point on the circle and whose legs cut the same circle is called inscribed angle.

If we mark the points in which the legs of the inscribed angle cut the circle with points A and B, then we call that angle the inscribed angle over the segment AB.

For these two points we can also define another angle whose legs also go through those points, but its vertex is placed in the center. This is called the central circle.

The central and inscribed angle theorem

The central angle over the arc of the circle is equal to the double inscribed angle over that same arc.

center inscribed angle theorem

Proof. There are exactly three different cases that may appear.

1. Center of the circle lies on one leg of the inscribed angle

center of circle lies on one leg of inscribed angle

In this case we can observe triangle TSB.

\mid TS \mid = \mid SB \mid = r

Which means that triangle TSB is isosceles triangle which implies that angles STB and SBT are equal. \alpha is the external angle of triangle TSB which means that he is equal to the sum of two opposite angles – STB and SBT.

This leads us to \alpha = 2 * \beta

2. Center of the circle lies within inscribed angle.

center lies within inscribed angle

For the second case we can divide these two angles in half. By doing so, we simplified the task and we can see that we only have to apply the first case two times.

\beta_1 = 2 * \alpha_1

\beta_2 = 2 * \alpha_2

\beta_1 + \beta_2 = 2(\alpha_1 + \alpha_2)

\beta = 2 * \alpha

3. The center lies outside of the inscribed circle

center lies outside of the inscribed circle

For proof of this case we’ll also use case 1.

\angle DSB = 2 \angle DTB

\angle DSA = 2 \angle DTA

 

When we add these two equations we get

\beta = 2 (\angle DTB + \angle DTA)

\beta = 2 * \alpha

 

 

 

Example 1. Find the missing angle:

find missing angle

\alpha = 2 * \beta = 2 * 50^{\circ} = 100^{\circ}

 

Example 2. Find the missing angle.

task find missing angle

\beta = \frac{1}{2} * \alpha

 

Thales’ theorem

Every inscribed angle over diameter of the circle is the right angle.

Proof. First, draw a circle and a triangle whose one side is the diameter and whose vertex is any point on the circle different from A and B.

Draw a line that connects C and O.

Triangle \triangle AOC is isosceles.

inscribed triangle with center\mid AO \mid = \mid OC \mid = R

\angle OAC = \angle OCA = \alpha

 

Triangle BOC is isosceles.

 

\mid AO \mid = \mid OC \mid = R

\angle OBC = \angle OCB = \beta

 

2 * \alpha + 2 * \beta = 180^{\circ} \rightarrow \alpha + \beta = 90^{\circ}

 

 

thales theorem

 

Reverse of Thales’ theorem

Center of the circumscribed circle of right triangle is located in the middle of hypotenuse.

reverse of thales theoremProof.

Divide the angle at point c with a line cx in a way that \angle ACX = \alpha.

 

Triangle ACX is isosceles \mid AX\mid = \mid CX \mid

\angle BCX = \angle CBX = \beta

Triangle CBX is isosceles!

\mid BX \mid = \mid CX \mid = \mid AX \mid

 

This means that x is the center of circumscribed circle and middle point of hypotenuse.

 

Every curve has its equation. From that equation we can find all the information we need to draw that curve and know every one of its properties.

Circles’ equation can be derived from its definition.

Let’s say that the coordinate of the circles center is S(p, q) and any point T has coordinate T(x, y). From the formula for the distance of two points and the fact that we know that that distance is equal to the length of the radius r we get that:

\sqrt{(x - p)^2 + (y - q)^2} = r

i.e.

\ (x - p)^2 + (y - q)^2 = r^2

If S(p, q) is the center of the circle, and r its radius, then the equation is \ (x - p)^2 + (y - q)^2 = r^2

circle equation

What if we put the center of the circle in origin and set the radius of 1? We get the unit circle.

Its equation is: x^2 + y^2 = 1

Example 1. Draw the circle that is described by the following equation.

(x - 2)^2 + (y + 2)^2 = 9

From the left side of this equation we can find the center of the circle. The x coordinate of the center is 2, and the y coordinate is – 2.

From the right side of the equation we can find radius. Radius is equal to \sqrt{9} = \pm 3. Since radius can only be a positive number we can see that the radius is equal to 3.

example of circle equation

Mutual position of line and circle

Line and circle can be in three positions.
1. Two different points – line is the secant of the circle.

line intersects circle in two points

2. One point in common – line is the tangent of the circle.

line and circle have one point - tangent

3. No points in common

line and circle have no points in common

How can you find out in which relation are the circle and the line without having to draw them? You can do that using their equations.

To find their relation we have to find their common points.

Example 1. In which relation are the line 2x – y + 3 = 0 and the circle x^2 + y^2 - 7x + 2y - 53 = 0?

First, from the line equation we extract y or x. in this example we’ll extract y.

y = 2x + 3

And then insert this y into the equation.

x^2 + (2x + 3)^2 - 7x + 2 (2x + 3) - 53 = 0

x^2 + 4x^2 + 12x + 9 - 7x + 4x + 6 - 53 = 0

5x^2 + 9x - 38 = 0

By solving this quadratic equation we get the solutions x_1 = 2, x_2 = \frac{16}{5}. By inserting them into any of these equations we get matching y coordinates and finally the points:

(2, 7), (\frac{16}{5}, \frac{23}{5}). This means that the given line and circle intersect in two points.

If we got only one point this would mean that the line is tangent of the circle and if the solutions were complex this would mean that the line and the circle have no points in common.

Circle power

Let k be a circle and T any point of the plane. For any line p which goes through point T, the product is a constant, where A and B are the intersections of the line p and circle k.

Proof.

1. T lies on the circle.

This means that T = A or T = B which means that either \mid TA \mid = 0 or \mid TB \mid = 0 \rightarrow \mid TA \mid * \mid TB \mid = 0

circle power proof 1

2. T lies within the circle.

\angle ACD = \angle ABD

\angle ATC = \angle DTB

This means that the triangles ACT and BTD are similar.

\frac{\mid TA \mid}{\mid TC \mid} = \frac{\mid TD \mid}{\mid TB \mid} \Rightarrow \mid TA \mid * \mid TB \mid = \mid TC \mid * \mid TD \mid

circle power proof 2

3. T lies outside of the circle.

First we’ll draw tangent on the circle from the point T.

Triangle SNA is congruent to triangle SNB – > \mid NA \mid = \mid NB \mid

circle power proof 3

\mid TA \mid * \mid TB \mid = (\mid TN \mid - \mid AN \mid) * (\mid TN \mid + \mid NB \mid) = \mid TN \mid^2 - \mid NA \mid^2 = \mid TN \mid^2 - (\mid SA \mid^2 - \mid NS \mid^2) =  \mid TN \mid^2 - \mid SA \mid^2 + \mid NS \mid^2 = \mid TS \mid^2 - \mid SC \mid^2 = \mid TC \mid^2

This means that this circle power does not depend on A and B.

Real number \mid TA \mid * \mid TB \mid is called circle power of the point T considering circle k.

Tangent and normal of the circle

The equation of the tangent

(x - p)^2 + (y + q)^2 = r^2

with touching point T_0 (x_0, y_0) is

(x_0 - p )(x - p) + (y_0 - q)(y - q) = r^2

Normal is a line that is perpendicular to the tangent and goes through the touching point of the tangent and the circle.

The equation of the normal

(x - p)^2 + (y + q)^2 = r^2

With touching point of the tangent and the circle T_0 (x_0, y_0) is

y - y_0 = \frac{y_0 - q}{x_0 - p} (x - x_0)

Example. Determine the equations of tangent and normal on the circle x^2 + y^2 + 8x - 6y - 1 = 0 in the point T with coordinates T (1, 4)

First thing we have to do is convert this equation to a form we are used to.

(x^2 + 8x) + (y^2 - 6y) - 1 = 0

(x + 4)^2 - 16 + (x - 3)^2 - 9 - 1 = 0

(x + 4)^2 + (x - 3)^2 = 26

Now we can clearly see everything we have to know if we want to find equations of normal and the tangent.

p = -4, q = 3, r = \sqrt{26}, x_0 = 1, y_0 = 4

For tangent:

(x_0 - p)(x - p) + (y_0 - q)(x - q) = r^2

(1 + 4)(x + 4) + (4 - 3)(x - 3) = 26

 

Circle worksheets

Graphing

  Constructing circles (174.0 KiB, 147 hits)

Naming

  Angles (375.8 KiB, 136 hits)

  Arcs (273.7 KiB, 121 hits)

Measuring

  Measurement of angles (455.2 KiB, 229 hits)

  Measurement of arcs (425.5 KiB, 142 hits)

  Measure the area and circumference (171.1 KiB, 147 hits)

  Measure the inscribed angles of a circle (735.9 KiB, 138 hits)

  Measure the angle between secants and tangents (710.6 KiB, 142 hits)

Calculation

  Tangents (589.7 KiB, 148 hits)

  Arc length (239.4 KiB, 146 hits)

  Sector area (239.6 KiB, 146 hits)

  Finding different segment measures - easy (424.1 KiB, 145 hits)

  Finding different segment measures - advanced (390.5 KiB, 135 hits)

  Make equations for the circle (670.4 KiB, 141 hits)

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