Tangents to Circles

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:

Find the tangents to a circle $\mathcal{C}$ of a given slope.

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):

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:

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 $y=-\frac43 x-\frac{26}{3}$ .

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:

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:

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:

$48m^2-5m+60$ 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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Tangents to Circles

Problem

Example

Solution (a) i:

Solution (a) (ii)

Solution (b) (i)

Solution (b) (ii)

Solution (b) (iii)

Remark

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J.P. McCarthy Maths

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Tangents to Circles

Problem

Example

Solution (a) i:

Solution (a) (ii)

Solution (b) (i)

Solution (b) (ii)

Solution (b) (iii)

Remark

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