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Lesson Explainer: Properties of Tangents and Chords Mathematics • 11th Grade

In this explainer, we will learn how to find the equation of a tangent to a circle and the equation of a perpendicular bisector of a chord using their perpendicularity.

Recall that a tangent to a circle is a line that meets the circle in exactly one point. The fundamental point about tangents to circles, which underlies much of what is to follow, is that at the point of intersection, a tangent is perpendicular to the radius of the circle.

Property: Tangents Are Perpendicular to Radii

If a line 𝐿 is tangent to a circle with center 𝐢 at a point 𝑃, then 𝐿 is perpendicular to the radius 𝐢𝑃.

Therefore, if we are given the coordinates of a circle’s center 𝐢(π‘₯,𝑦) and are told that a line 𝐿 is tangent to the circle at a point 𝑃(π‘₯,𝑦), then we can write down the slope of the radius 𝐢𝑃 as slopeofchangeinchangein𝐢𝑃=𝑦π‘₯=π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯.

With tangent 𝐿 and radius 𝐢𝑃 being perpendicular, the slope of the tangent is the negative reciprocal of the slope of the radius. That is, slopeof𝐿=βˆ’π‘₯βˆ’π‘₯π‘¦βˆ’π‘¦.

We now know the slope of 𝐿 and that 𝐿 passes through the point 𝑃(π‘₯,𝑦). This is enough information to write down an equation for 𝐿: 𝑦=π‘šπ‘₯+𝑏=βˆ’π‘₯βˆ’π‘₯π‘¦βˆ’π‘¦π‘₯+𝑏, where the constant 𝑏 is determined by substituting in the coordinates of 𝑃, 𝑦=π‘¦οŒ― and π‘₯=π‘₯: 𝑦=βˆ’π‘₯βˆ’π‘₯π‘¦βˆ’π‘¦π‘₯+𝑏𝑦=βˆ’π‘₯βˆ’π‘₯π‘¦βˆ’π‘¦π‘₯+𝑏𝑏=𝑦+π‘₯βˆ’π‘₯π‘¦βˆ’π‘¦π‘₯.

Let us try an example.

Example 1: Finding the Equation of a Tangent to a Circle at a Given Point

Determine the equation of the tangent to the circle (π‘₯+4)+(π‘¦βˆ’5)=34 at the point 𝐷(1,2).

Answer

Since we are given the equation of the circle in standard form, we may read off the coordinates of its center directly as (βˆ’4,5). The slope of the radius at 𝐷 is therefore slopeofradiuschangeinchangein=𝑦π‘₯=2βˆ’51βˆ’(βˆ’4)=βˆ’35=βˆ’35.

Since we know that the tangent at a point is always perpendicular to the radius at that point, we know that the slope of the tangent is the negative reciprocal of the slope of the radius. That is, slopeoftangentπ‘š=βˆ’ο€Όβˆ’53=53.

We now know the slope of the tangent and that it passes through the point 𝐷(1,2). This is enough information to write down an equation for the tangent: 𝑦=π‘šπ‘₯+𝑏=53π‘₯+𝑏, with constant 𝑏 to be determined by substituting the coordinates of 𝐷: 𝑦=53π‘₯+𝑏2=53Γ—1+𝑏𝑏=2βˆ’53=13.

Thus, the equation of the tangent line at the point 𝐷 is given by 𝑦=53π‘₯+13.

In the previous example, we were given the coordinates of a point on a circle where a line is tangent. We were able to use perpendicularity to calculate the equation of the tangent line. Let us now look at an example in which the slope of the tangent is given and we use perpendicularity to find the coordinates of points of intersection.

Example 2: Finding Two Possible Tangents to a Circle given the Gradient of the Tangent

The circle (π‘₯βˆ’4)+(π‘¦βˆ’4)=10 lies between two lines of slope βˆ’13 such that both lines are tangent to the circle. Write down the equations of the two lines in the form 𝑦=π‘šπ‘₯+𝑐.

Answer

A quick sketch of the setup looks as follows.

Since we are given the slopes of the tangent lines, the information we need to be able to write down their equations is the coordinates of some points on the lines. To find these points, we will use the fact that there is a perpendicular radius at the point where a tangent meets a circle. The line containing the radii we are interested in is therefore the line passing through the center of the circle, with a slope equal to the negative reciprocal of the slope of the tangent.

So, we proceed in four steps:

  1. Identify the coordinates of the center 𝐢 of the circle.
  2. Find the equation of the line 𝐿 passing through 𝐢 and is perpendicular to the tangent lines.
  3. Find the coordinates of the points π‘ƒοŠ§ and π‘ƒοŠ¨ where 𝐿 intersects the circle.
  4. Use the coordinates of the points π‘ƒοŠ§ and π‘ƒοŠ¨ to find the equations of the tangents π‘‡οŠ§ and π‘‡οŠ¨.

By inspecting the equation of the circle (π‘₯βˆ’4)+(π‘¦βˆ’4)=10, we see that the coordinates of the center are (4,4).

We are told that the slope of the tangent is βˆ’13. The slope of the line 𝐿 containing the perpendicular radii is therefore the negative reciprocal of this, or 3.

We can write down an equation for the line 𝐿: 𝑦=3π‘₯+𝑏 and calculate the value of the constant 𝑏 by substituting in the coordinates of the circle center, π‘₯=4 and 𝑦=4: 4=3Γ—4+𝑏𝑏=βˆ’8.

We have πΏβˆΆπ‘¦=3π‘₯βˆ’8.

The coordinates of the points π‘ƒοŠ§ and π‘ƒοŠ¨ where the line 𝐿 (and the two tangent lines) intersects the circle satisfy the equation of the line 𝑦=3π‘₯βˆ’8 and the circle (π‘₯βˆ’4)+(π‘¦βˆ’4)=10. We can therefore substitute 3π‘₯βˆ’8 for 𝑦 in (π‘₯βˆ’4)+(π‘¦βˆ’4)=10 and simplify as follows: (π‘₯βˆ’4)+(3π‘₯βˆ’8βˆ’4)=10π‘₯βˆ’8π‘₯+16+9π‘₯βˆ’72π‘₯+144=1010π‘₯βˆ’80π‘₯+150=0.

The roots of this quadratic 10ο€Ήπ‘₯βˆ’8π‘₯+15ο…οŠ¨ are the π‘₯-coordinates of the intersection points. Factoring 10ο€Ήπ‘₯βˆ’8π‘₯+15=10(π‘₯βˆ’3)(π‘₯βˆ’5), we find that these π‘₯-coordinates are π‘₯=3 and π‘₯=5. We can now substitute these π‘₯-values back into 𝑦=3π‘₯βˆ’8 to find the 𝑦-coordinates of the intersection points: 𝑦=3π‘₯βˆ’8=3Γ—3βˆ’8=1 and 𝑦=3π‘₯βˆ’8=3Γ—5βˆ’8=7.

Finally, we can use the coordinates of these two points to calculate the equations of the tangent lines, where 𝑦=βˆ’13π‘₯+𝑏1=βˆ’13Γ—3+𝑏𝑏=1+1=2 gives us 𝑦=βˆ’13π‘₯+2 as the equation of the first and 𝑦=βˆ’13π‘₯+𝑏7=βˆ’13Γ—5+𝑏𝑏=7+53=263 gives us 𝑦=βˆ’13π‘₯+263 as the equation of the second.

In the previous example, we made essential use of the line 𝐿 being perpendicular to the two tangent lines. Notice that whenever we have a pair of parallel lines tangent to a circle, there is a unique line 𝐿 perpendicular to them passing through the center of the circle. This line defines a diameter of the circle.

Having investigated tangent lines, we turn our attention now to chords and in particular, their perpendicular bisectors. A chord is simply a line segment connecting any two points on the circumference of a circle. Given a chord, we can construct its perpendicular bisector by finding the midpoint of the chord and then computing the equation of the perpendicular line passing through that point.

Example 3: Finding the Perpendicular Bisector of a Chord

  1. Consider the circle π‘₯+𝑦+16π‘₯+16𝑦+115=0 and the line 5𝑦+π‘₯+35=0 intersecting it at the points 𝑃 and 𝑄. Find the coordinates of the points 𝑃 and 𝑄.
  2. Find the equation of the perpendicular bisector of the chord 𝑃𝑄 in the form 𝑦=π‘Žπ‘₯+𝑏, where π‘Ž and 𝑏 are constants.

Answer

Part 1

To find the coordinates of the points 𝑃 and 𝑄, we use the fact that these points lie on the line 5𝑦+π‘₯+35=0 and on the circle π‘₯+𝑦+16π‘₯+16𝑦+115=0 and so their coordinates satisfy both equations. This means that we can rearrange the equation of the line to find an expression for either π‘₯ or 𝑦 and substitute this into the equation of the circle. Since the coefficient of π‘₯ is 1 in 5𝑦+π‘₯+35=0, it will probably be easier to work with π‘₯. So, let us rearrange 5𝑦+π‘₯+35=0π‘₯=βˆ’5π‘¦βˆ’35 and then substitute this into the equation of the circle: π‘₯+𝑦+16π‘₯+16𝑦+115=0(βˆ’5π‘¦βˆ’35)+𝑦+16(βˆ’5π‘¦βˆ’35)+16𝑦+115=025𝑦+350𝑦+1225+π‘¦βˆ’80π‘¦βˆ’560+16𝑦+115=026𝑦+286𝑦+780=026𝑦+11𝑦+30=0.

This yields a quadratic equation 26𝑦+11𝑦+30=26(𝑦+6)(𝑦+5)=0 whose roots 𝑦=βˆ’5 and 𝑦=βˆ’6 are the 𝑦-coordinates of the points where the line intersects the circle. Substituting these values back into the equation of the line gives us the corresponding π‘₯-coordinates: π‘₯=βˆ’5π‘¦βˆ’35=βˆ’5Γ—βˆ’5βˆ’35=βˆ’10 and π‘₯=βˆ’5π‘¦βˆ’35=βˆ’5Γ—βˆ’6βˆ’35=βˆ’5.

Hence, we have found the coordinates 𝑃(βˆ’10,βˆ’5) and 𝑄(βˆ’5,βˆ’6).

Part 2

For the second part of the question, we need to find the equation of the perpendicular bisector of the chord 𝑃𝑄. Recall that the perpendicular bisector is the unique line perpendicular to 𝑃𝑄 passing through its midpoint.

The coordinates of the midpoint are given by 𝑀=ο€½βˆ’10+(βˆ’5)2,βˆ’5+(βˆ’6)2=ο€Όβˆ’152,βˆ’112.

We now have a point lying on the perpendicular bisector: π‘€ο€Όβˆ’152,βˆ’112. Next, we need to calculate its slope. Since the perpendicular bisector is perpendicular to 𝑃𝑄, its slope is the negative reciprocal of the slope of 𝑃𝑄. We can work out the slope of 𝑃𝑄 using the endpoints 𝑃(βˆ’10,βˆ’5) and 𝑄(βˆ’5,βˆ’6), which we calculated in part one of the question: slopeof𝑃𝑄=π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯=βˆ’6βˆ’(βˆ’5)βˆ’5βˆ’(βˆ’10)=βˆ’15.

The slope of the perpendicular bisector is the negative reciprocal of βˆ’15, which is to say 5, and we use the coordinates ο€Όβˆ’152,βˆ’112 to find its equation: 𝑦=5π‘₯+π‘βˆ’112=5Γ—ο€Όβˆ’152+𝑏𝑏=βˆ’112βˆ’5Γ—ο€Όβˆ’152=32.

The equation of the perpendicular bisector of the chord 𝑃𝑄 can be given by 𝑦=5π‘₯+32.

Our interest in perpendicular bisectors comes down to the following fact: the perpendicular bisector of a chord always passes through the center of the circle.

Theorem: Perpendicular Bisectors of Chords

Let 𝐢 be a circle and 𝑃𝑄 be a chord. Then, the perpendicular bisector of 𝑃𝑄 passes through the center of 𝐢.

Proof:

Write 𝑂 for the center of the circle. Consider the triangle obtained by joining the endpoints 𝑃 and 𝑄 to the center 𝑂 by radii.

The line segments 𝑂𝑃 and 𝑂𝑄 are the same length because they are radii. That is, 𝑂𝑃𝑄 is an isosceles triangle. Now, drop a perpendicular from the center 𝑂 to the chord 𝑃𝑄. This line is perpendicular to 𝑃𝑄 by construction, and since it divides the triangle 𝑂𝑃𝑄 into two congruent (right) triangles (by the RHS congruence criterion), it must bisect 𝑃𝑄, which is to say that it is the unique perpendicular bisector of 𝑃𝑄.

We can use this fact to solve various problems involving circles and chords.

Example 4: Finding the Missing Coordinate of the Center of a Circle given Two Endpoints of a Chord

The points 𝑃(2,βˆ’3) and 𝑄(8,βˆ’1) lie on the circle with center 𝐢(π‘˜,βˆ’5), where π‘˜ is a constant. Work out the value of π‘˜ by considering the perpendicular bisector of the chord 𝑃𝑄.

Answer

We want to find the π‘₯-coordinate π‘˜ of the center of the circle, having been given its 𝑦-coordinate of βˆ’5. We are told to consider the perpendicular bisector of the chord 𝑃𝑄. As we have the coordinates of 𝑃 and 𝑄, we can find the equation of the perpendicular bisector by considering the slope and midpoint of 𝑃𝑄.

We know that the perpendicular bisector of any chord in a circle passes through the center of the circle, so we will be able to find the π‘₯-coordinate of the circle’s center by finding the point with a 𝑦-coordinate of βˆ’5 on the perpendicular bisector.

So, the first thing to do is to write down an equation for the perpendicular bisector. This we do using the slope of the chord 𝑃𝑄, slopeof𝑃𝑄=π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯=βˆ’1βˆ’(βˆ’3)8βˆ’2=26=13, and the coordinates of its midpoint, given by 𝑀=ο€½2+82,βˆ’3+(βˆ’1)2=(5,βˆ’2).

The perpendicular bisector has a slope equal to the negative reciprocal of 13, that is, βˆ’3, and passes through (5,βˆ’2). Substituting these coordinates, 𝑦=βˆ’3π‘₯+π‘βˆ’2=βˆ’3Γ—5+𝑏𝑏=13.

Thus, we obtain the equation 𝑦=βˆ’3π‘₯+13. Because it is the perpendicular bisector of a chord, we know that the center 𝐢(π‘˜,βˆ’5) lies on this line. Therefore, we can substitute again to find π‘˜: 𝑦=βˆ’3π‘₯+13βˆ’5=βˆ’3Γ—π‘˜+13βˆ’18=βˆ’3Γ—π‘˜π‘˜=6.

We finish with an example of finding tangents to a circle, which are parallel to a chord specified by its endpoints.

Example 5: Finding the Equation of a Tangent Parallel to a Chord

The points 𝑃(1,1) and 𝑄(4,2) lie on the circle (π‘₯βˆ’2)+(π‘¦βˆ’3)=5. Find the equations of the tangents parallel to the chord 𝑃𝑄, giving your answers in the form 𝑦=π‘šπ‘₯+𝑏 accurate to two decimal places.

Answer

We are trying to find a pair of tangents to the circle.

These tangents are parallel to the chord 𝑃𝑄 and are therefore perpendicular to the diameter of the circle defined by the perpendicular bisector of 𝑃𝑄. The intersection points 𝐴 and 𝐡 of this perpendicular bisector with the circle are the points where the corresponding tangents meet the circle.

That is to say, the coordinates of 𝐴 and 𝐡 satisfy both the equation of the perpendicular bisector and the equation of the circle. This will allow us to calculate these coordinates by substitution.

Since the two tangent lines are parallel to the chord 𝑃𝑄, we know that their slopes are equal to the slope of 𝑃𝑄. This fact, along with the coordinates of 𝐴 and 𝐡 that we have calculated, is enough information to write down equations of the tangent lines.

First, we write down an equation for the perpendicular bisector of 𝑃𝑄 by calculating the slope of 𝑃𝑄: slopeof𝑃𝑄=2βˆ’14βˆ’1=13 and the coordinates of its midpoint: 𝑀=ο€Ό1+42,2+12=ο€Ό52,32.

We flip the slope 13 and take its negative to get the slope of the perpendicular bisector: βˆ’3. Now, we substitute the slope and coordinates into the point–slope form of the equation of a straight line, (π‘¦βˆ’π‘¦)=π‘š(π‘₯βˆ’π‘₯)π‘¦βˆ’32=βˆ’3ο€Όπ‘₯βˆ’52, and simplify and rearrange to make 𝑦 the subject, 𝑦=βˆ’3π‘₯+152+32=βˆ’3π‘₯+9.

The equation of the perpendicular bisector of 𝑃𝑄 is 𝑦=βˆ’3π‘₯+9. We now want to calculate the coordinates of the points 𝐴 and 𝐡 where this line intersects the circle. We know that the coordinates of these points satisfy both 𝑦=βˆ’3π‘₯+9 and the equation of the circle. We therefore substitute βˆ’3π‘₯+9 for 𝑦 in (π‘₯βˆ’2)+(π‘¦βˆ’3)=5 to find the points 𝐴 and 𝐡: it is at these points that the two tangents meet the circle: (π‘₯βˆ’2)+((βˆ’3π‘₯+9)βˆ’3)=5π‘₯βˆ’4π‘₯+4+9π‘₯βˆ’36π‘₯+36=510π‘₯βˆ’40π‘₯+35=05ο€Ή2π‘₯βˆ’8π‘₯+7=0.

We can use the quadratic formula to calculate the roots of this equation: π‘₯=2Β±12√2. These are the π‘₯-coordinates of the intersection points. Now, we substitute back into 𝑦=βˆ’3π‘₯+9 to get the 𝑦-coordinates: 𝑦=3βˆ“32√2. We therefore have coordinates 𝐴2βˆ’12√2,3+32√2 and 𝐡2+12√2,3βˆ’32√2.

Since the tangents we are looking for are parallel to the chord 𝑃𝑄 of slope 13, they must also have slope 13. We therefore have one tangent, 𝑦=13π‘₯+𝑏, passing through the point 𝐴2βˆ’12√2,3+32√2. We can substitute in the π‘₯- and 𝑦-coordinates of the point 𝐴 to find the constant π‘οŠ§: 3+32√2=13ο€Ό2βˆ’12√2+𝑏𝑏=3+32√2βˆ’23+16√2=73+53√2, which evaluates to 𝑏=4.69 to two decimal places. This gives us an equation for one tangent: 𝑦=0.33π‘₯+4.69.

The other tangent has the equation 𝑦=13π‘₯+π‘οŠ¨ and passes through the point 𝐡2+12√2,3βˆ’32√2. Substituting as before, we have 3βˆ’32√2=13ο€Ό2+12√2+𝑏𝑏=3βˆ’32√2βˆ’23βˆ’16√2=73βˆ’53√2, which evaluates to 𝑏=βˆ’0.02 to two decimal places. This gives us an equation for the other tangent: 𝑦=0.33π‘₯βˆ’0.02.

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • We know that a tangent to a circle at a point is perpendicular to the radius at that point. We can use this fact to write down the equation of a tangent given any point on a circle.
  • Given the slope of a tangent to a circle and the equation of the circle, we can work out the coordinates of the point where the tangent meets the circle by constructing a perpendicular line passing through the center of the circle.
  • We know that the perpendicular bisector of a chord passes through the midpoint of the circle. We can use this fact to calculate the equations of tangent lines parallel to given chords.

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