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Circle

The definition and basics of a circle 

Let S be a fixed point in the plane and r>0. A circle is a set of all points of a plane which are equidistant from point S which is equal to r. Point S is called the center of the circle. The distance r from the center to any point of the circle is called the radius of the circle. A circle with a center situated in point S and with radius r is denoted by c(S,r).

 

a circle with center S and radius r

A chord is the line segment in which the ends are two different points that lie on the circumference of a circle. A chord is denoted by c.

 

a chord

 

The longest chord is called the diameter. A diameter contains the center of the circle. The length of a diameter is equal to 2r. A diameter is denoted by R.

diameter

 

A sector is an enclosed part of a circle which contains two radii and a part of a circumference.

 

a sector

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

a segment of a circle

 

A secant of a circle is the line that intersects a circle in two points and a tangent line to the circle is a line that has exactly one point in common with the circle. A secant is denoted by s and a tangent by t.

A secant and tangent to the circle

Let t be the tangent to the circle c(S,|SD|) and a tangent with the circle has one in common point D. Point D is called a point of tangency and is valid: \overline{SD} \perp t.

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, consider circle on which lie two different points A and B. Those points divide it in two different arcs. Their union is the whole circle.

For denoting arcs we must refer to the positive direction – in an anticlockwise motion.

This means that \widehat{AB} \not= \widehat{BA}.

We can draw an angle whose axis is the center of the circle and whose legs intersect the circle at points A and B. This angle formed is called the central angle.

A whole circle has a circumference of 360^{\circ}. This means that we can make the following ratio:

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

i.e.

    \[l ( 1^{\circ}) = \frac{r\pi}{180^{\circ}},\]

where l is length of the arc.

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

    \[l (\alpha) = \frac{r\pi\alpha}{180^{\circ}}.\]

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

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

and

    \[P (\alpha) = \frac{r^2 \pi\alpha}{360^{\circ}}.\]

 

Central and inscribed angles

Every angle whose vertex is a point on the circle and whose legs intersect the circle is called an inscribed angle.

If we mark the points in which the legs of the inscribed angle intersect the circle with  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 run through those points, but its vertex is placed in the center. This is called the central angle.

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.

 

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

1.) The center of the circle lies on one leg of the inscribed angle.

 

In this case we can observe triangle TSB. We have:

    \[\mid TS \mid = \mid SB \mid = r,\]

which means that triangle TSB is an isosceles triangle which implies STB=SBT=\beta. \alpha is the external angle of triangle TSB which means that the measure of angle \alpha is equal to the sum of measures of angles  STB and SBT, which is equal to 2\beta.

 

2.) The center of the circle lies within inscribed angle.

First of all, determine point C in the arc \widehat{AB} which is diametrically opposite of point T. We then apply case 1.) on inscribed angles \angle{ATS} and \angle{BTS}:

\angle{ASC}+\angle{BSC}=2\angle{ATS}+2\angle{BTS}=2(\angle{ATS}+\angle{BTS}).

Considering that

\beta = \angle{ATS}+ \angle{BTS} and \alpha =\angle{ASC}+\angle{BSC},  we have:

\alpha = 2 \beta.

3.) The center lies outside of the inscribed circle.

To prove  this case we will also use case 1.).  First, we have that \alpha =\angle{ASB}= \angle{ASC} - \angle{CSB} and \beta = \angle{ATB}=\angle{ATC} - \angle{BTC}. Then we have:

\angle{ASC} = 2 \angle{ATC},

\angle{BSC} = 2 \angle{BTC}.

.When we subtract  these two equations we get:

\angle{ASC} - \angle{BSC} = 2 (\angle{ATC} -\angle{BTC} ).

It follows:

\alpha = 2\beta.

Example 1. Find the missing angle:

 

Solution:

\alpha = 2 \beta = 2 \cdot 50^{\circ} = 100^{\circ}.

 

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 C on the circle different from A and B.

Draw a segment line that connects point C and the center of the circle S.

 

Triangle BSC is an isosceles triangle because |BS|=|CS| (length of \overline{BS} and \overline{CS} is equal to the radius of the circle). It follows \angle SBC = \angle SCB = \alpha.

For the same reason triangle ASC is an isosceles triangle and \angle SAC = \angle SCA = \beta.

Now, we have:

2 \alpha + 2 \beta = 180^{\circ} \Rightarrow \alpha + \beta = 90^{\circ}.

 

Reverse of Thales’ theorem

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

Proof.

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

Triangle ACX is an isosceles triangle, it follows \mid AX\mid = \mid CX \mid and \angle ACX = \angle CAX = \alpha.

Also, triangle CBX is an isosceles triangle, it follows \mid BX\mid = \mid CX \mid and \angle BCX = \angle CBX = \beta.

Thus we obtain \mid BX \mid = \mid CX \mid = \mid AX \mid.

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

 

Equation of a circle

Every curve has an equation. From this equation we can find all the  information we require to draw the curve and know each one of its properties.

A circle’s 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 on the circumference of a circle has the coordinate T(x, y). From the formula for the distance of two points and the fact that we know that the 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 of a circle 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 a unit circle.

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

Example 2. Draw a circle that is described by the following equation:

    \[(x - 2)^2 + (y + 2)^2 = 9.\]

Solution:

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 a radius can only be a positive number we can see that the radius is equal to 3.

 

The mutual position of  a line and a circle

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

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

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

Solution:

First, from the line equation we extract y or x. in this example we will extract y:

y = 2x + 3.

And then insert this y into the equation.

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

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

\Rightarrow  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 a tangent to the circle and if the solutions were complex this would mean that the line and the circle have no points in common.

Power of the point considering circle

Let c(S,r) 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 c. I.e.:

    \[|TA|\cdot|TB|=|ST|^{2} - r^{2}.\]

Real number |TA| \cdot |TB| is called power of the point T considering circle c.

Depending on whether the point T is located; T is on the circumference of the circle c(S, r), T lies within the circle c(S, r) or T is outside of the circle c(S, r), the power of the point T relative to the circle is equal to 0, negative or positive.

A tangent and normal to the circle

The equation of a tangent of the circle (x - p)^2 + (y + q)^2 = r^2 with touching point T_0 (x_0, y_0) is:

    \[(x_0 - p )(x - p) + (y_0 - q)(y - q) = r^2.\]


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

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

Solution:

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

    \[(x^2 + 8x) + (y^2 - 6y) - 1 = 0\Rightarrow (x + 4)^2 - 16 + (x - 3)^2 - 9 - 1 = 0 \Rightarrow (x + 4)^2 + (x - 3)^2 = 26. \]

 

From the equation of a circle we can see that the center of the circle is located in point S( -4, 3).

Points S and T determine a normal to the circle, so we get that

    \[y-3=\frac{4-3}{-4-1} (x+4),\]

i.e.

    \[y=-\frac{1}{5} x + \frac{11}{5}\]

is the required equation of a normal to the circle (x + 4)^2 + (x - 3)^2 = 26 in point T. The equation of a tangent to the circle we will base using the perpendicularity clause of two lines. We get that

    \[y=5x + \frac {11}{5}\]

is the required equation of a tangent to the circle (x + 4)^2 + (x - 3)^2 = 26 in point T.

Circle worksheets

Graphing

  Constructing circles (174.0 KiB, 288 hits)

Naming

  Angles (375.8 KiB, 269 hits)

  Arcs (273.7 KiB, 265 hits)

Measuring

  Measurement of angles (455.2 KiB, 492 hits)

  Measurement of arcs (425.5 KiB, 291 hits)

  Measure the area and circumference (171.1 KiB, 248 hits)

  Measure the inscribed angles of a circle (735.9 KiB, 257 hits)

  Measure the angle between secants and tangents (710.6 KiB, 261 hits)

Calculation

  Tangents (589.7 KiB, 294 hits)

  Arc length (239.4 KiB, 298 hits)

  Sector area (239.6 KiB, 303 hits)

  Finding different segment measures - easy (424.1 KiB, 290 hits)

  Finding different segment measures - advanced (390.5 KiB, 236 hits)

  Make equations for the circle (670.4 KiB, 292 hits)

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