As I said in the previous post, there is a duality:

Points on a Curve (Geometry) Solutions of an Equation (Algebra)

This means we can answer geometric questions using algebra and answer algebraic questions using geometry.

### Problem

Consider the following two questions:

Find the tangents to a circle of a given slope.Find the tangents to a circle through a given point.

Both can be answered using the duality principle.

#### Example

*Find the tangents to the circle*

*that are*

(a)

parallel to the line(b)

through the point[caution: the numbers here are disgusting]

*Solution (a) i:*

* *First of all a sketch (and the remark that a tangent is a line):

*Here we see the circle and the line on the bottom left. The two tangents we are looking for are as shown. They have the same slope as and have only one intersection with . These two pieces of information will allow us to find the equations of the tangents.*

Let us first find the slope of by writing it in the form:

.

.

Now any line parallel to also has slope and thus is of the form:

,

for some (the ‘lower/bottom’ line will have the smaller -value).

Now the second condition is that a tangent to a circle intersects it at a single point only. Now an intersection between two curves is a point on both curves and by duality the coordinates of the intersection satisfy both curve’s equations *at the same time. *That is we want the following simultaneous equations to have a single solution only (in general an has two):

Now this is a quadratic in . It is a quadratic and so has geometry. For it to have a single root it must look like:

*A quadratic has a single (repeated) root when .*

Therefore the circle and line of slope has a unique solution if:

Note that this is a quadratic in — and so has two solutions as expected. Multiplying out and simplifying gives:

.

The ‘‘ formula can be used to solve this however by multiplying both sides by three and dividing both sides before we have:

.

Looking at the factors of we rewrite the middle term

,

and therefore the two tangents are:

and .

*The two tangents, along with which actually intersects the curve twice (so that the simultaneous equations of and would have two solutions and so ).*

##### Solution (a) (ii)

The preceding solution is nice and sound. However, if we take a theorem on board we get a nicer solution:

*The tangent to a circle is perpendicular to the diameter. This means that the perpendicular distance from the centre to a tangent is equal to the radius…*

The perpendicular distance, , from a line (and this is when it is necessary to use this format rather than the much more useful ) to a point is given by:

.

Now lines parallel to are of the form … and using , the centre of is at . Therefore, for to be a tangent to , it must be the case that the perpendicular distance to the centre

,

must equal to the radius

so that

.

It is possible to square both sides but easier just to say so that we have

or

or ,

so that the two tangents have equations:

and .

**Exercise: **Are these the same solutions as before?

##### Solution (b) (i)

First of all, a picture:

*In this case we know a point on the line but not the slope, .*

We know that the point is on these lines but we don’t know the slopes, . Therefore the tangents are of the form:

.

We want these lines to be such that the intersection between them and is unique. Therefore we substitute into the equation of :

.

Multiplying out and tidying up:

.

Now, if this is to have a single solution, we once again require :

.

Multiplying out an rearranging we get:

.

Using the ‘‘ formula we find the two values of and thus tangents:

.

##### Solution (b) (ii)

Another option is to use geometry in the first instance to find the following circle:

*The red circle here cuts at the contact points and .*

Where is the centre of the circle, we can find the distance () using the distance formula:

.

We know the radius, the distance . We know that so that is a right-angled-triangle satisfying Pythagoras:

.

Therefore, the red circle has centre and radius $\sqrt{123}$ so equation:

.

The contact points and $latex $B$ are the points where these circles intersect… so we find the intersection of the two circles using simultaneous equations:

.

.

Subtract one from the other:

.

Now back-substitute:

.

Substitute this into to find the contact points:

Therefore we have two pairs of points each giving a tangent.

**Exercise (messy): **Show that this gives the same answer as above.

##### Solution (b) (iii)

A third solution goes as follows. Write the two tangents in the form and then . Now there are two unknowns here but we have two pieces of information:

- the point is on the tangent
- the perpendicular distance from the tangent to the centre is equal to five.

Write down these to generate simultaneous equations:

.

**Exercise:** Substitute into the second equation.** **Take separately. Alternatively square both sides of the second to get:

.

Solve for and then . Confirm you have same answer as above.

##### Remark

A very interesting thing happens with, say, which lies inside the circle.

Then, without drawing a picture, we might say that lines through have form:

.

Trying to find the intersection of this line and the circle we get:

.

When we expand this out and tidy this:

.

If this is to have one solution we must have . Implementing this leads onto:

,

which has no real solutions:

* has no real roots… there is no tangent from $latex to the circle… of course tangents to circles always lie *outside* the circle and so can’t contain internal points.*

## Leave a comment

Comments feed for this article