# 9.6: Tangent half-lines

Suppose (ABC) is an arc of a circle (Gamma). A half-line ([AX)) is called tangent to the arc (ABC) at (A) if the line ((AX)) is tangent to (Gamma), and the points (X) and (B) lie on the same side of the line ((AC)).

If the arc is formed by the line segment ([AC]) then the half-line ([AC)) is considered to be tangent at (A). If the arc is formed by a union of two half lines ([AX)) and ([BY)) in ((AC)), then the half-line ([AX)) is considered to be tangent to the arc at (A).

Proposition (PageIndex{1})

The half-line ([AX)) is tangent to the arc (ABC) if and only if

(measuredangle ABC + measuredangle CAX equiv pi).

Proof

For a degenerate arc (ABC), the statement is evident. Further we assume the arc (ABC) is nondegenerate.

Applying Theorem 9.1.1 and Theorem 9.2.1, we get that

(2 cdot measuredangle ABC + 2 cdot measuredangle CAX equiv 0).

Therefore, either

(measuredangle ABC + measuredangle CAX equiv pi), or (measuredangle ABC + measuredangle CAX equiv 0).

Since ([AX)) is the tangent half-line to the arc (ABC, X) and (B) lie on the same side of ((AC)). By Corollary 3.4.1 and Theorem 3.3.1, the angles (CAX), (CAB), and (ABC) have the same sign. In particular, (measuredangle ABC + measuredangle CAX otequiv 0); that is, we are left with the case

(measuredangle ABC +measuredangle CAX equiv pi).

Exercise (PageIndex{1})

Show that there is a unique arc with endpoints at the given points (A) and (C), that is tangent to the given half line ([AX)) at (A).

Hint

If (C in (AX)), then the arc is the line segment ([AC]) or the union of two half-lines in ((AX)) with vertices at (A) and (C).

Assume (C otin (AX)). Let (ell) be the perpendicular line dropped from (A) to ((AX)) and (m) be the perpendicular bisector of ([AC]).

Note that (ell parallel m); set (O = ell cap m). Note that the circle with center (O) passing thru (A) is also passing thru (C) and tangent to ((AX)).

Note that one the two arcs with endpoints (A) and (C) is tangent to ([AX)).

The uniqueness follow from Proposition (PageIndex{1}).

Exercise (PageIndex{2})

Let ([AX)) be the tangent half-line to an arc (ABC). Assume (Y) is a point on the arc (ABC) that is distinct from (A). Show that (measuredangle XAY o 0) as (AY o 0).

Hint

Use Proposition (PageIndex{1}) and Theorem 7.4.1 to show that (measuredangle XAY = measuredangle ACY). By Axiom IIIc, (measuredangle ACY o 0) as (AY o 0); hence the result.

Exercise (PageIndex{3})

Given two circle arcs (AB_1C) and (AB_2C), let ([AX_1)) and ([AX_2)) at (A), and ([CY_1)) and ([CY_2)) be the half-lines tangent to the arcs (AB_1C) and (AB_2C) at (C). Show that

(measuredangle X_1AX_2 equiv -measuredangle Y_1CY_2.)

Hint

Apply Proposition (PageIndex{1}) twice.

(Alternatively, apply Corollary 5.4.1 for the reflection across the perpendicular bisector of ([AC]).)

## Finding, without derivatives, the line through $(9,6.125)$ that is tangent to the parabola $y=-frac18x^2+8$

What part I am looking at right now is f). My parabola equation for the front edge of the roof is $y=-frac18x^2+8$ . All the laser lights have an equation that is tangent to the parabola except the red laser beam. The point that I have for the red laser beam source’s location for the tower is at $(9,6.125)$ . How can I determine an equation of the lines that are tangent to the parabola and pass through $(9,6.125)$ , without using derivatives?

I tried to find an equation that goes through the vertex and got the equation -5/24x+8, then found the midpoint of the x on the two intersections, (0+5/3)/2= 5/6. After that, I plugged it into the equation and recieved 2279/288. I thought the line passing through (5/6, 2279/288) and (9,6.125) would be tangent to the parabola. Unfortunately this was incorrect, and I do not know how to solve this without derivatives.

## Circles: Diameter, Chord, Radius, Arc, Tangent

In these lessons, we will learn the following parts of a circle: diameter, chord, radius, arc and tangent.

We will also learn about congruent circles, concentric circles and intersecting circles.

The following figures show the different parts of a circle: tangent, chord, radius, diameter, minor arc, major arc, minor segment, major segment, minor sector, major sector. Scroll down the page for more examples and explanations.

### Circle

In geometry, a circle is a closed curve formed by a set of points on a plane that are the same distance from its center O. That distance is known as the radius of the circle.

### Diameter

The diameter of a circle is a line segment that passes through the center of the circle and has its endpoints on the circle. All the diameters of the same circle have the same length.

### Chord

A chord is a line segment with both endpoints on the circle. The diameter is a special chord that passes through the center of the circle. The diameter would be the longest chord in the circle.

The radius of the circle is a line segment from the center of the circle to a point on the circle. The plural of radius is radii.

In the above diagram, O is the center of the circle and and are radii of the circle. The radii of a circle are all the same length. The radius is half the length of the diameter.

An arc is a part of a circle.

In the diagram above, the part of the circle from B to C forms an arc.

An arc can be measured in degrees.

In the circle above, arc BC is equal to the ∠BOC that is 45°.

### Tangent

A tangent is a line that touches a circle at only one point. A tangent is perpendicular to the radius at the point of contact. The point of tangency is where a tangent line touches the circle.

In the above diagram, the line containing the points B and C is a tangent to the circle.

It touches the circle at point B and is perpendicular to the radius . Point B is called the point of tangency.

### Parts Of A Circle

The following video gives the definitions of a circle, a radius, a chord, a diameter, secant, secant line, tangent, congruent circles, concentric circles, and intersecting circles.

A secant line intersects the circle in two points.

A tangent is a line that intersects the circle at one point.

A point of tangency is where a tangent line touches or intersects the circle.

Congruent circles are circles that have the same radius but different centers.

Concentric circles are two circles that have the same center, but a different radii.

Intersecting Circles: Two circles may intersect at two points or at one point. If they intersect at one point then they can either be externally tangent or internally tangent.

Two circles that do not intersect can either have a common external tangent or common internal tangent.
In the common external tangent, the tangent does not cross between the two circles.
In the common internal tangent, the tangent crosses between the two circles.

### Parts Of A Circle, Including Radius, Chord, Diameter, Central Angle, Arc, And Sector

Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

## What is a Tangent?

A tangent is a line (or line segment) that intersects a circle at exactly one point. To do that, the tangent must also be at a right angle to a radius (or diameter) that intersects that same point.

In our crop circle U, if we look carefully, we can see a tangent line off to the right, line segment FO. That would be the tiny trail the circlemakers walked along to get to the spot in the field where they started forming their crop circle. Crop circles almost always "appear" very close to roads and show some signs of tangents, which is why most researchers say they are made by human pranksters.

The word "tangent" comes from a Latin term meaning "to touch," because a tangent just barely touches a circle. Tangents, of course, also allude to writing or speaking that diverges from the topic, as when a writer goes off on a tangent and points out that most farmers do not like having their crops stomped down by vandals from this or any other world.

## Dielectric Constant, Strength, & Loss Tangent

Values presented here are relative dielectric constants (relative permittivities). As indicated by er = 1.00000 for a vacuum, all values are relative to a vacuum.

Multiply by &epsilon0 = 8.8542 x 10 -12 F/m (permittivity of free space) to obtain absolute permittivity. Dielectric constant is a measure of the charge retention capacity of a medium.

In general, low dielectric constants (i.e., Polypropylene) result in a "fast" substrate while large dielectric constants (i.e., Alumina) result in a "slow" substrate.

The dielectric loss tangent is defined by the angle between the capacitor's impedance vector and the negative reactive axis, as illustrated in the diagram to the right. It determines the lossiness of the medium. Similar to dielectric constant, low loss tangents result in a "fast" substrate while large loss tangents result in a "slow" substrate.

Beware that the exact values can vary greatly depending on the particular manufacturer's process, so you should seek out data from the manufacturer for critical applications.

The dielectric constant can be calculated using: ε = Cs / Cv , where Cs is the capacitance with the specimen as the dielectric, and Cv is the capacitance with a vacuum as the dielectric.

The dissipation factor can be calculated using: D = tan δ = cot θ = 1 / (2 π f RpCp) , where δ is the loss angle, θ is the phase angle, f is the frequency, Rp is the equivalent parallel resistance, and Cp is the equivalent parallel capacitance.

Note: All values can vary by very large amounts depending on the specific material. Check with the MatWeb.com website for more details. Other resources: Electrical Properties of Insulators, Dielectric Properties of Materials.

Supplemental information provided by website visitor James S. for complex dielectric:

The dielectric constants at the top of [this] page are reminiscent of the propagation constants given by Roald K. Wangsness, Electromagnetic Fields, 2nd Ed., John Wiley & Sons, New York, 1986, p. 383, Eq. (24-42) and (24-43). The sixth equation given on the web page is correct. That equation, as given by P. Hoekstra and A. Delaney in Dielectric properties of soils at UHF and microwave frequencies, J. Geophys. Res., v. 79, 10 Apr 1974, p. 1699, ". is written as

K ' (&omega) is the dielectric constant and

K " (&omega) is the dielectric loss factor.

Note: Thanks to Gareth for correcting the omission of a square root sign in the dielectric equations.
Thanks to Craig B. for correcting the loss tangent for Teflon (0.00028 rather than 0.0028).

Please Support RF Cafe by purchasing my ridiculously low−priced products, all of which I created.

## Characterization and bifurcation diagram of the family of quadratic differential systems with an invariant ellipse in terms of invariant polynomials

Consider the class QS of all non-degenerate planar quadratic systems and its subclass QSE of all its systems possessing an invariant ellipse. This is an interesting family because on one side it is defined by an algebraic geometric property and on the other, it is a family where limit cycles occur. Note that each quadratic differential system can be identified with a point of (<>>^<12>) through its coefficients. In this paper we provide necessary and sufficient conditions for a system in QS to have at least one invariant ellipse. We give the global “bifurcation” diagram of the family QS which indicates where an ellipse is present or absent and in case it is present, the diagram indicates if the ellipse is or it is not a limit cycle. The diagram is expressed in terms of affine invariant polynomials and it is done in the 12-dimensional space of parameters. This diagram is also an algorithm for determining for any quadratic system if it possesses an invariant ellipse and whether or not this ellipse is a limit cycle.

This is a preview of subscription content, access via your institution.

## Side Length of Tangent & Secant of a Circle

If a secant and a tangent of a circle are drawn from a point outside the circle, then the product of the lengths of the secant and its external segment equals the square of the length of the tangent segment.

### Practice Problems

Use the theorem for the intersection of a tangent and a secant of a circle to solve the problems below.

##### Problem 1

In this diagram, the red line is a tangent, how long is it?

$ed x^2 = (7 + lue 5) cdot lue 5 ed x^2 = ( 12 ) cdot lue 5 ed x^2 = 60 ed x = sqrt <60>$

##### Problem 2

In the problem below, the red line is a tangent of the circle, what is its length?

$ed x^2 = (9+ lue 7) cdot lue 7 ed x^2 = (16) cdot lue 7 ed x^2 = 112 ed x = sqrt < 112 >$

#### Two Secants Intersecting

If two secant segments are drawn from a point outisde a circle, the product of the lengths (C + D) of one secant segment and its exteranal segment (D) equals the product of the lengths (A + B) of the other secant segment and its external segment (B).

##### Problem 3

Use the theorem above to determine A if $B = 4, C = 8, D = 5$ .

$(A + lue 4) cdot lue 4 = (8 + lue 5) cdot lue 5 (A + lue 4) cdot lue 4 = (14 ) cdot lue 5 (A + lue 4) cdot lue 4 = 65 frac 1 4 cdot (A + lue 4) cdot lue 4 =frac 1 4 cdot 65 frac <1> < cancel 4>cdot (A + lue 4) cdot lue > = 16.25 A + lue 4 - 4 = 16.25 - 4 A = 12.25$

##### Problem 4

Use the theorem above to determine A if $B = 8, C = 16, D = 10$ .

$(A + lue 8) cdot lue 8 = (16 + lue 10) cdot lue 10 (A + lue 8) cdot lue 8 = (26) cdot lue 10 (A + lue 8) cdot lue 8 = 260 frac1 8 cdot (A + lue 8) cdot lue 8 = frac1 8 cdot 260 (A + lue 8) = 32.5 A = 32.5 - 8 A = 24.5$

##### Problem 5

The two secants in the picture below are not drawn to scale. If $KO = 16$, $KJ = 4$, and $LO = 32$ , what is $LM$ ?

The first challenge here is for you to recognize that the side lengths we are given are not the ones that we can use for the formula. Please see the diagram below, which labels the measurements that we are given.:

What we need to know is the length of $overline$ and $MO$ !

So, the first thing we must do is is figure out the portion of the secants that are outside the circle.

$color<#666600> < JO >= 16 - 4 = 12 ext KO cdot JO = LO cdot MO 16 cdot 12 = 32 cdot MO 192 = 32 cdot MO frac< 192> <32>= frac<32> <32>cdot MO 6 = MO ext LM = LO-MO LM = 32 - 6 = oxed < 26>$

KO &bull JO = LO &bull MO
--> JO = KO &minus KJ
--> JO = 16 &minus 4 = 12
--> KO &bull JO = LO &bull MO
--> MO
--> MO
--> MO
--> MO
--> LM = LO &minus MO
--> LM = 32 &minus 6 =26 -->

If the equation of $C_2$ is $(x-h)^2+(y-k)^2=r^2$

Again, the gradient of $C_2$ at $(9,6)$

$=-dfrac<9-h><6-k>$ which should be $=$ the gradient of the parabola at $(9,6)$ $dfrac<4><2cdot6>$

Find the equation of both the circles using $S+kL=0$ .

$L$ is the equation of the tangent to the parabola at points of contact. Point of contact may be considered as point circles.

Since point of contact comes out to be $(4,4)$ and $(9,6)$ and the equation of tangents are: $2y=x+4$ and $3y=x+9$

Circle through $(9,6)$ is $(x-9)^2+(y-6)^2+k(3y-x-9)=0$

Since it passes through the focus $(1,0)$ , therefore, putting the values of $X$ and $Y$ we get $K=10$ .

Hence the equation comes out to be: $(x-9)^2+(y-6)^2+10(3y-x-9)=0$ or $x^2+y^2-28x+18y+108=0$

Similarly, for the second circle through $(4,4)$ : $(x-4)^2+(y-4)^2+k(2y-x-4)=0$

Since it passes through $(1,0)$ we get $K$ as $5$ . We have got both the circles and we can find the radius.

Using this method we get the centre of circles as $(13/2,-1)$ and $(14,-9)$ .

Therefore using the centres and the point $(1,0)$ we can get the radius using the distance formula.

$r_1$ comes out to be $sqrt<125/4>$ and $r_2$ comes out to be $sqrt<250>$ .

## Tangents and Slopes

We&rsquoll use three relations we already have. First, tan A = sin A / cos A. Second, sin A = a/c. Third, cos A = b/c. Dividing a/c by b/c and canceling the c&rsquos that appear, we conclude that tan A = a/b. That means that the tangent is the opposite side divided by the adjacent side:

#### Slopes of lines

The point B is where the line cuts the y-axis. We can let the coordinates of B be (0,b) so that b, called the y-intercept, indicates how far above the x-axis B lies. (This notation conflicts with labeling the sides of a triangle a, b, and c, so let&rsquos not label the sides right now.)

You can see that the point 1 unit to the right of the origin is labeled 1, and its coordinates, of course, are (1,0). Let C be the point where that verical line cuts the horizontal line through B. Then C has coordinates (1,b).

The point A is where the vertical line above 1 cuts the original line. Let m denote the distance that A is above C. Then A has coordinates (1,b+m). This value m is called the slope of the line. If you move right one unit anywhere along the line, then you&rsquoll move up m units.

Now consider the angle CBA. Let&rsquos call it the angle of slope. It&rsquos tangent is CA/BC = m/1 = m. Therefore, the slope is the tangent of the angle of slope.

#### Angles of elevation and depression

The term &ldquoangle of elevation&rdquo refers to the angle above the horizontal from the viewer. If you&rsquore at point A, and AH is a horizontal line, then the angle of elevation to a point B above the horizon is the angle BAH. Likewise, the &ldquoangle of depression&rdquo to a point C below the horizon is the angle CAH.

Tangents are frequently used to solve problems involving angles of elevation and depression.

#### Common angles again

Note that the tangent of a right angle is listed as infinity. That&rsquos because as the angle grows toward 90°, it&rsquos tangent grows without bound. It may be better to say that the tangent of 90° is undefined since, using the circle definition, the ray out from the origin at 90° never meets the tangent line.

90°&pi/201infinity
0100

#### Exercises

29. In a right triangle a = 30 yards and tan A = 2. Find b and c.

49. cos t = 2 tan t. Find the value of sin t.

Note: In the following problems distance means horizontal distance, unless otherwise stated the height of an object means its height above the horizontal plane through the point of observation. The height of the observer&rsquos eye is not to be taken into account unless specially mentioned. In problems involving the shadow of an object the shadow is supposed to fall on the horizontal plane through the base of the object, unless otherwise stated.

151. The angle of elevation of a tree 250 feet distant is 16° 13'. Find the height.

152. Find the height of a steeple distant 321 feet, angle of elevation 35 ° 16'.

153. From a ship, the angle of elevation of the top of a lighthouse 200 feet above the water is 2° 20'. Find the distance.

154. From the top of a lighthouse 165 feet above the water the angle of depression of a ship is 3° 50'. Find the distance.

159. Find the height of a tower, distant 186 feet, angle of elevation 40° 44'.

160. On one side of a stream a pole 50 feet high has from an opposite point an angle of elevation of 5° 33'. Find the breadth of the stream.

164. From one hill the top of another 128 feet higher has an angle of elevation of 2° 40'. Find the distance.

165. From one hill to the top of another distant 6290 feet has an angle of elevation of 4° 9'. Find how much the height of the second hill exceeds that of the first.

189. The gable end of a roof measures 40 feet across at the base and 26 feet from the base to the ridge. At what angle do the rafters slope?

#### Hints

29. Since you know a and tan A, you can find b. You can then determine c by the Pythagorean theorem, or by using sines, or by using cosines.

49. You&rsquoll need two identities. First, tan t = sin t/cos t. Second, the Pythagorean identity, sin 2 t + cos 2 t = 1. Then you&rsquoll have to solve a quadratic equation.

151. Remember that the tangent of an angle in a right triangle is the opposite side divided by the adjacent side. You know the adjacent side (the distance to the tree), and you know the angle (the angle of elevation), so you can use tangents to find the height of the tree.

152. You know the angle (again, the angle of elevation) and the adjacent side (the distance to the steeple), so use tangents to find the opposite side.

153. Using the angle and the opposite side, use tangent to find the adjacent side.

154. Same hint as in 153.

159. Same hint as in 152.

160. Same hint as in 153.

164. Same hint as in 153.

165. Same hint as in 152.

189. The gable end of a roof is an isosceles triangle. If you drop a perpendicular line from the ridge, you get two congruent right triangles. You know the two legs of the triangles, so you can determine the angle of slope of the rafters using arctangent.

29. b = a/tan A = 30/2 = 15 yards. c = 33.5 yards.

49. Since cos t = 2 tan t, therefore cos t = 2 sin t/cos t, so cos 2 t = 2 sin t, and, by the Pythagorean identity, you get 1 &ndash sin 2 t = 2 sin t. That gives you a quadratic equation sin 2 t + 2 sin t &ndash 1 = 0. The solutions are sin t = &ndash1 ± &radic2. Of these two solutions the only feasible one is sin t = &radic2 &ndash 1.

151. Height = 250 tan 16°13' = 72.7' = 72'9".

152. Height = 321 tan 35°16' = 227 feet.

153. Distance = 200/tan 2°20' = 4908 feet, nearly a mile.

154. Distance = 165/tan 3°50' = 2462 feet, nearly half a mile.

159. Height = 186 tan 40°44' = 160 feet.

160. Distance = 50/tan 5°33' = 515 feet.

164. Distance = 128/tan 2°40', about 2750 feet, a little more than half a mile.