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 is tangent to a circle with center at a point , then is perpendicular to the radius .

Therefore, if we are given the coordinates of a circle’s center and are told that a line is tangent to the circle at a point , then we can write down the slope of the radius as

With tangent and radius being perpendicular, the slope of the tangent is the negative reciprocal of the slope of the radius. That is,

We now know the slope of and that passes through the point . This is enough information to write down an equation for : where the constant is determined by substituting in the coordinates of , and :

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 at the point .

### Answer

Since we are given the equation of the circle in standard form, we may read off the coordinates of its center directly as . The slope of the radius at is therefore

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,

We now know the slope of the tangent and that it passes through the point . This is enough information to write down an equation for the tangent: with constant to be determined by substituting the coordinates of :

Thus, the equation of the tangent line at the point is given by .

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 lies between two lines of slope such that both lines are tangent to the circle. Write down the equations of the two lines in the form .

### 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:

- Identify the coordinates of the center of the circle.
- Find the equation of the line passing through and is perpendicular to the tangent lines.
- Find the coordinates of the points and where intersects the circle.
- Use the coordinates of the points and to find the equations of the tangents and .

By inspecting the equation of the circle , we see that the coordinates of the center are .

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

We can write down an equation for the line : and calculate the value of the constant by substituting in the coordinates of the circle center, and :

We have .

The coordinates of the points and where the line (and the two tangent lines) intersects the circle satisfy the equation of the line and the circle . We can therefore substitute for in and simplify as follows:

The roots of this quadratic are the -coordinates of the intersection points. Factoring , we find that these -coordinates are and . We can now substitute these -values back into to find the -coordinates of the intersection points: and

Finally, we can use the coordinates of these two points to calculate the equations of the tangent lines, where gives us as the equation of the first and gives us as the equation of the second.

In the previous example, we made essential use of the line 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 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

- Consider the circle and the line intersecting it at the points and . Find the coordinates of the points and .
- Find the equation of the perpendicular bisector of the chord in the form , where and are constants.

### Answer

**Part 1**

To find the coordinates of the points and , we use the fact that these points lie on the line and on the circle and so their coordinates satisfy both equations. This means that we can rearrange the equation of the line to find an expression for either or and substitute this into the equation of the circle. Since the coefficient of is 1 in , it will probably be easier to work with . So, let us rearrange and then substitute this into the equation of the circle:

This yields a quadratic equation whose roots and are the -coordinates of the points where the line intersects the circle. Substituting these values back into the equation of the line gives us the corresponding -coordinates: and

Hence, we have found the coordinates and .

**Part 2**

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

The coordinates of the midpoint are given by

We now have a point lying on the perpendicular bisector: . Next, we need to calculate its slope. Since the perpendicular bisector is perpendicular to , its slope is the negative reciprocal of the slope of . We can work out the slope of using the endpoints and , which we calculated in part one of the question:

The slope of the perpendicular bisector is the negative reciprocal of , which is to say 5, and we use the coordinates to find its equation:

The equation of the perpendicular bisector of the chord can be given by .

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

**Proof:**

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

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

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 and lie on the circle with center , where is a constant. Work out the value of by considering the perpendicular bisector of the chord .

### Answer

We want to find the -coordinate of the center of the circle, having been given its -coordinate of . We are told to consider the perpendicular bisector of the chord . As we have the coordinates of and , we can find the equation of the perpendicular bisector by considering the slope and midpoint of .

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 -coordinate of the circle’s center by finding the point with a -coordinate of 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 , and the coordinates of its midpoint, given by

The perpendicular bisector has a slope equal to the negative reciprocal of , that is, , and passes through . Substituting these coordinates,

Thus, we obtain the equation . Because it is the perpendicular bisector of a chord, we know that the center lies on this line. Therefore, we can substitute again to find :

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 and lie on the circle . Find the equations of the tangents parallel to the chord , giving your answers in the form 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 and are therefore perpendicular to the diameter of the circle defined by the perpendicular bisector of . The intersection points and of this perpendicular bisector with the circle are the points where the corresponding tangents meet the circle.

That is to say, the coordinates of and 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 , we know that their slopes are equal to the slope of . This fact, along with the coordinates of and 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 by calculating the slope of : and the coordinates of its midpoint:

We flip the slope and take its negative to get the slope of the perpendicular bisector: . Now, we substitute the slope and coordinates into the point–slope form of the equation of a straight line, and simplify and rearrange to make the subject,

The equation of the perpendicular bisector of is . We now want to calculate the coordinates of the points and where this line intersects the circle. We know that the coordinates of these points satisfy both and the equation of the circle. We therefore substitute for in to find the points and : it is at these points that the two tangents meet the circle:

We can use the quadratic formula to calculate the roots of this equation: . These are the -coordinates of the intersection points. Now, we substitute back into to get the -coordinates: . We therefore have coordinates and .

Since the tangents we are looking for are parallel to the chord of slope , they must also have slope . We therefore have one tangent, passing through the point . We can substitute in the - and -coordinates of the point to find the constant : which evaluates to to two decimal places. This gives us an equation for one tangent:

The other tangent has the equation and passes through the point . Substituting as before, we have which evaluates to to two decimal places. This gives us an equation for the other tangent:

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.