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

Points on a Curve (Geometry) \Leftrightarrow 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:

  1. Find the tangents to a circle \mathcal{C} of a given slope.
  2. Find the tangents to a circle \mathcal{C} through a given point.

Both can be answered using the duality principle.

Example

Find the tangents to the circle

\mathcal{C}\equiv x^2+y^2-4x+6y-12=0

that are

(a) parallel to the line L\equiv 4x+3y+20=0

(b) through the point (-10,-5) [caution: the numbers here are disgusting]

Solution (a) i:

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

circle.jpg

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

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

y=mx+c.

3y=-4x-20

\displaystyle\Rightarrow y=-\frac43 x -\frac{20}{3}.

Now any line parallel to L also has slope m=-\frac43 and thus is of the form:

\displaystyle y=-\frac{4}{3}x+c,

for some c (the ‘lower/bottom’ line will have the smaller c-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 x^2 has two):

x^2+y^2-4x+6y-12=0

\displaystyle y=-\frac{4}{3}x+c

\displaystyle \Rightarrow x^2+(c-\frac43 x)^2-4x+6\left(c-\frac43x\right)-12=0

\displaystyle \Rightarrow x^2+c^2-\frac83 cx+\frac{16}{9}x^2-4x+6c-8x-12=0

\displaystyle \left[1+\frac{16}{9}\right]x^2+\left[-\frac83 x-12\right]c+(c^2+6c-12)=0

\displaystyle \frac{25}{9}x^2+(-12-\frac83 c)x+(c^2+6c-12)=0

Now this is a quadratic in x. It is a +x^2 quadratic and so has \bigcup geometry. For it to have a single root it must look like:

quad.jpg

A quadratic has a single (repeated) root when b^2-4ac=0.

Therefore the circle and line of slope -4/3 has a unique solution if:

\displaystyle \left(-12-\frac83 c\right)^2-4\left(\frac{25}{9}\right)(c^2-6c-12)=0

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

\displaystyle 144+\frac{192}{3}c+\frac{64}{9}c^2-\frac{100}{9}c^2-\frac{200}{3}c+\frac{400}{3}=0

\displaystyle \Rightarrow 4c^2-\frac{8}{3}c-\frac{832}{3}=0.

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

3c^2+2c-208=0.

Looking at the factors of 3\times 208=624 we rewrite the middle term

3c^2-24c+26c-208=0

\Rightarrow 3c(c-8)+26(c-8)=0

\Rightarrow (c-8)(3c+26)=0

\displaystyle \Rightarrow c=8\text{ or }-\frac{26}{3},

and therefore the two tangents are:

\displaystyle y=-\frac43 x+8 and y=-\frac43 x-\frac{26}{3}.

graph1.jpg

The two tangents, along with L which actually intersects the curve twice (so that the simultaneous equations of \mathcal{C} and L would have two solutions and so b^2-4ac>0).

Solution (a) (ii)

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

circle2.jpg

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, d, from a line ax+by+c=0 (and this is when it is necessary to use this format rather than the much more useful y=mx+c) to a point (x_1,y_1) is given by:

\displaystyle d=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}.

Now lines parallel to 4x+3y+20=0 are of the form L_{\lambda}\equiv 4x+3y+\lambda =0… and using (-g,-f), the centre of \mathcal{C} is at (2,-3). Therefore, for L_{\lambda} to be a tangent to \mathcal{C}, it must be the case that the perpendicular distance to the centre (2,-3)

\displaystyle d=\frac{|4(2)+3(-3)+\lambda|}{\sqrt{4^2+3^2}},

must equal to the radius

r=\sqrt{g^2+f^2-c}=\sqrt{(-2)^2+3^2-(-12)}=\sqrt{25}=5 so that

\frac{|\lambda-1|}{5}=5

\Rightarrow |\lambda -1|=25.

It is possible to square both sides but easier just to say |a|=\pm a so that we have

\lambda-1=25 or \lambda-1=-25

\Rightarrow \lambda =26 or -24,

so that the two tangents have equations:

4x+3y+26=0 and 4x+3y-24=0.

Exercise: Are these the same solutions as before?

Solution (b) (i)

First of all, a picture:

circle3.jpg

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

We know that the point (-10,-5) is on these lines but we don’t know the slopes, m. Therefore the tangents are of the form:

y+5=m(x+10)\Rightarrow y=mx+10m-5.

We want these lines to be such that the intersection between them and \mathcal{C} is unique. Therefore we substitute y=mx+10m-5 into the equation of \mathcal{C}:

x^2+(mx+10m-5)^2-4x+6(mx+10m-5)-12=0.

Multiplying out and tidying up:

 (m^2+1)x^2+(20m^2-4m-4)x+(100m^2-40m-17)=0.

Now, if this is to have a single solution, we once again require b^2-4ac=0:

(20m^2-4m-4)^2-4(m^2+1)(100m^2-40m-17)=0.

Multiplying out an rearranging we get:

476m^2-192m-84=0

\Rightarrow 119m^2-48m-21=0.

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

\displaystyle y=\left(\frac{24}{119}\pm \frac{5}{119}\sqrt{123}\right)(x+10)-5.

graph2

Solution (b) (ii)

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

circle4

The red circle here cuts \mathcal{C} at the contact points A and B.

Where c(2.-3) is the centre of the circle, we can find the distance |cP| (P(-10,-5)) using the distance formula:

|cP|=\sqrt{(2+10)^2+(-3+5)^2}=\sqrt{148}=2\sqrt{37}.

We know the radius, the distance |Ac|=|Bc|=5. We know that cA\perp cP so that \Delta  AcP is a right-angled-triangle satisfying Pythagoras:

|cP|^2=|cA|^2+|AP|^2=\Rightarrow 148=25+|Ap|^2\Rightarrow |Ap|=\sqrt{123}.

Therefore, the red circle has centre (-10,-5) and radius $\sqrt{123}$ so equation:

(x+10)^2+(y+5)^2=123

\Rightarrow x^2+y^2+20x+10y+2=0.

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

x^2+y^2+20x+10y+2=0.

x^2+y^2-4x+6y-12=0.

Subtract one from the other:

24x+4y+14=0

\displaystyle \Rightarrow y=-6x-\frac72.

Now back-substitute:

\displaystyle x^2+\left(-6x-\frac72\right)^2+20x+10\left(-6x-\frac72\right)+2=0

\displaystyle \Rightarrow 37x^2+2x-\frac{83}{4}=0

\displaystyle \Rightarrow x=-\frac{1}{37}\pm\frac{5}{74}\sqrt{123}.

Substitute this into \displaystyle y=-6x-\frac72 to find the contact points:

\displaystyle \left(-\frac{1}{37}\pm\frac{5}{74}\sqrt{123},-\frac{247}{37}\mp \frac{15}{37}\sqrt{123}\right)

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 y=mx+c and then mx-y+c=0. Now there are two unknowns here but we have two pieces of information:

  • the point (-10,-5) is on the tangent
  • the perpendicular distance from the tangent to the centre (2,-3) is equal to five.

Write down these to generate simultaneous equations:

c=10m-5

|2m+3+c|=5\sqrt{m^2+1}.

Exercise: Substitute c=10m-5 into the second equation. Take 12m-2=\pm 5\sqrt{m^2+1} separately. Alternatively square both sides of the second to get:

(12m-2)^2=25(m^2+1).

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

Remark

A very interesting thing happens with, say, (5,-2) which lies inside the circle.

Then, without drawing a picture, we might say that lines through (5,-2) have form:

y=m(x-5)-2.

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

x^2+(m(x-5)+2)^2-4x+6(m(x-5)+2)-12=0.

When we expand this out and tidy this:

(m^2+1)x^2+x(2m-10m^2-4)+(25m^2-10m-20)=0.

If this is to have one solution we must have b^2-4ac=0. Implementing this leads onto:

48m^2-5m+60=0,

which has no real solutions:

graph3.jpg

48m^2-5m+60 has no real roots… there is no tangent from $latex (5,-2) to the circle… of course tangents to circles always lie outside the circle and so can’t contain internal points.

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