Coordinate Geometry: Straight Lines and Circles

Two circles lie entirely outside each other, so they share no point at all. Subtract one equation from the other. What is the line you get?

Take $S_1:x^2+y^2-1=0$ and $S_2:x^2+y^2-10x+21=0$. Their centres are $(0,0)$ and $(5,0)$ with radii $1$ and $2$, and since $5>1+2$ the circles are separate. Yet

$S_1-S_2=10x-22=0\Rightarrow x=2.2,$

a perfectly ordinary vertical line. It cannot be a common chord, because there is no common chord.

It is the radical axis: the locus of points from which the tangent lengths to the two circles are equal. From the point $(2.2,y)$ the two tangent lengths are

$\sqrt{(2.2{)}^2+y^2-1}\text{and}\sqrt{(2.2{)}^2+y^2-10(2.2)+21},$

and both equal $\sqrt{3.84+y^2}$. The subtraction $S_1-S_2=0$ always produces this locus; when the circles happen to meet, the locus passes through both intersections and is therefore also the common chord, which is why the formula is usually taught under that name.

The general lesson is one that runs through the chapter. An algebraic combination of two curve equations is itself a curve with a geometric meaning, and knowing that meaning converts long calculations into one line.

1. The forms of a line, and the two Advanced actually uses

Beyond the familiar slope and intercept forms, two others carry most of the harder questions.

The normal form $x\cos \alpha +y\sin \alpha =p$ describes a line by the length $p$ of the perpendicular from the origin and the angle $\alpha$ that perpendicular makes with the $x$-axis. It is the natural form whenever a distance from the origin is prescribed.

The parametric form is more useful still. Any point on the line through $(x_1,y_1)$ at angle $\theta$ is

$x=x_1+r\cos \theta ,y=y_1+r\sin \theta ,$

where $r$ is the signed distance from $(x_1,y_1)$. Substituting this into the equation of any second curve produces an equation in $r$ whose roots are the distances to the intersection points, so a question about distances along a line becomes a question about the roots of a quadratic.

A shift of origin is the other quiet simplifier. Replacing $x$ by $X+h$ and $y$ by $Y+k$ moves the origin to $(h,k)$ without rotating anything, so slopes, distances and angles are all unchanged while constant terms usually disappear. Choosing the new origin at a centre of symmetry, or at the intersection of two given lines, is often the difference between a page of algebra and three lines.

Illustration 1

A line through $(1,2)$ makes an angle of $45^{\circ }$ with the $x$-axis and meets the circle $x^2+y^2=25$ at $A$ and $B$. Find $PA\cdot PB$, where $P=(1,2)$.

Substituting $x=1+\frac{r}{\sqrt{2}}$ and $y=2+\frac{r}{\sqrt{2}}$ into the circle,

${\left(1+\frac{r}{\sqrt{2}}\right)}^2+{\left(2+\frac{r}{\sqrt{2}}\right)}^2=25\Rightarrow r^2+3\sqrt{2}r-20=0.$

The product of the roots is $-20$, so $PA\cdot PB=20$ in magnitude. The sign records that $P$ lies inside the circle, so the two intersections are on opposite sides — information the answer $20$ alone would have lost.

2. Angle bisectors: choosing the right one

Two intersecting lines have two bisectors, perpendicular to each other, and questions almost always want a specific one. Both are given by

$\frac{a_{1}x+b_{1}y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm \frac{a_{2}x+b_{2}y+c_2}{\sqrt{a_2^2+b_2^2}},$

and the choice of sign is what has to be made deliberately.

To find the bisector of the angle containing the origin, first arrange both equations so that $c_1$ and $c_2$ are positive, then take the $+$ sign. To decide which bisector splits the acute angle, compute $a_1{a}_2+b_1{b}_2$ after that same normalisation: if it is negative, the $+$ sign gives the acute bisector, and if positive, the $-$ sign does.

Illustration 2

Find the bisector of the angle between $3x-4y+7=0$ and $12x+5y-2=0$ that contains the origin.

Make both constants positive: the first already is, and the second becomes $-12x-5y+2=0$. Then take the $+$ sign:

$\frac{3x-4y+7}{5}=\frac{-12x-5y+2}{13}\Rightarrow 99x-27y+81=0,$

that is $11x-3y+9=0$. Substituting the origin into both original expressions gives $+7$ and $-2$, confirming that the origin lies in the region where the two normalised expressions have opposite signs, exactly as the construction assumed.

3. Families through an intersection

If $L_1=0$ and $L_2=0$ meet, then $L_1+\lambda L_2=0$ is a line through their intersection for every $\lambda$, and every such line arises this way except $L_2$ itself. This removes the need to find the intersection at all: impose one further condition and solve for $\lambda$.

The same idea covers concurrency. Three lines are concurrent exactly when the determinant of their coefficients vanishes, which is the algebraic statement that one of them is a combination of the other two.

Illustration 3

Find the line through the intersection of $x+2y-3=0$ and $3x-y+1=0$ that is perpendicular to $2x+y=5$.

Write the family as $\left(x+2y-3\right)+\lambda \left(3x-y+1\right)=0$, that is $(1+3\lambda )x+(2-\lambda )y+(\lambda -3)=0$. Its slope is $-\frac{1+3\lambda }{2-\lambda }$, and perpendicularity to a line of slope $-2$ requires this slope to be $\frac{1}{2}$:

$-\frac{1+3\lambda }{2-\lambda }=\frac{1}{2}\Rightarrow -2-6\lambda =2-\lambda \Rightarrow \lambda =-\frac{4}{5}.$

Substituting gives $x-14y+19=0$ up to a constant multiple, and checking that this line is perpendicular to $2x+y=5$ confirms the arithmetic.

4. Feet, images and distances

The foot of the perpendicular from $(x_1,y_1)$ to $ax+by+c=0$, and the image of that point in the line, both come from one relation:

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=-\frac{a{x}_1+b{y}_1+c}{a^2+b^2},$

which gives the foot; doubling the right-hand side gives the image, since the foot is the midpoint.

Illustration 4

Find the image of $(3,8)$ in the line $x+3y=7$.

Here $a{x}_1+b{y}_1+c=3+24-7=20$ and $a^2+b^2=10$, so for the image the common ratio is $-2\cdot \frac{20}{10}=-4$. Then

$x=3+1(-4)=-1,y=8+3(-4)=-4,$

giving $(-1,-4)$. Checking, the midpoint $(1,2)$ satisfies $1+6=7$, and the segment's direction $(-4,-12)$ is parallel to the normal $(1,3)$.

5. The four centres of a triangle

Each centre is the meeting point of a different family of lines, and each has its own formula worth carrying.

The centroid is the average of the vertices, $\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$, and it divides every median in the ratio $2:1$ from the vertex. The incentre is a weighted average, with each vertex weighted by the length of the side opposite it:

$I=\left(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c}\right).$

The circumcentre is equidistant from the three vertices and is found as the intersection of two perpendicular bisectors, while the orthocentre is the intersection of two altitudes. For a right-angled triangle both are immediate: the circumcentre is the midpoint of the hypotenuse and the orthocentre is the vertex containing the right angle.

The three that are not the incentre are always collinear, on the Euler line, with the centroid dividing the segment from orthocentre to circumcentre in the ratio $2:1$. In coordinates that reads $H=3G-2O$, which turns any two of them into the third without further work.

Illustration 5

Find the incentre of the triangle with vertices $A(0,0)$, $B(3,0)$ and $C(0,4)$.

The side opposite $A$ has length $a=BC=5$, opposite $B$ is $b=CA=4$, and opposite $C$ is $c=AB=3$. So

$I=\frac{5(0,0)+4(3,0)+3(0,4)}{12}=\frac{(12,12)}{12}=(1,1).$

Checking independently, the triangle is right-angled at the origin with inradius $\frac{b+c-a}{2}=1$, so the incentre must sit one unit from each leg, at $(1,1)$.

Illustration 6

A triangle has circumcentre $(1,2)$ and centroid $(2,3)$. Find its orthocentre.

Using $H=3G-2O$,

$H=3(2,3)-2(1,2)=(6-2,9-4)=(4,5).$

No vertex was needed. Questions supplying two centres and asking for the third are testing exactly this relation, and attempting to reconstruct the triangle first wastes the whole question.

6. The circle, its tangents, and the length of a tangent

A circle $x^2+y^2+2gx+2fy+c=0$ has centre $(-g,-f)$ and radius $\sqrt{g^2+f^2-c}$, which is real only when $g^2+f^2>c$. Writing $S$ for the left-hand side, the two most useful facts are that $S_1=S(x_1,y_1)$ is the square of the tangent length from an external point, and that the tangent at a point on the circle is $T=0$, where $T$ is $S$ with $x^2\to x{x}_1$, $y^2\to y{y}_1$, $2x\to x+x_1$ and $2y\to y+y_1$.

A line touches a circle exactly when the distance from the centre equals the radius. For $y=mx+k$ and $x^2+y^2=a^2$ this reduces to $k^2=a^2\left(1+m^2\right)$, so the tangents of a given slope come in a pair.

Illustration 7

Find the locus of a point from which the two tangents to $x^2+y^2=a^2$ are perpendicular.

Let the point be $(h,k)$. The tangents of slope $m$ through it satisfy $k=mh\pm a\sqrt{1+m^2}$, and squaring gives

$\left(h^2-a^2\right)m^2-2hkm+\left(k^2-a^2\right)=0.$

The two slopes are the roots, and perpendicularity means their product is $-1$:

$\frac{k^2-a^2}{h^2-a^2}=-1\Rightarrow h^2+k^2=2a^2.$

The locus is a concentric circle of radius $a\sqrt{2}$, called the director circle. The route — write the tangency condition as a quadratic in the slope, then use the relation between its roots — handles every "angle between the tangents" question.

7. Two circles

Everything about a pair of circles is decided by one number: the distance $d$ between their centres, compared with $r_1+r_2$ and $\left|r_1-r_2\right|$.

Two circles cut orthogonally when the tangents at a point of intersection are perpendicular, which by Pythagoras means $d^2=r_1^2+r_2^2$ and reduces, in general form, to $2g_1{g}_2+2f_1{f}_2=c_1+c_2$.

Illustration 8

Find the length of the common chord of $x^2+y^2=16$ and $x^2+y^2-6x+8=0$.

Subtracting gives the common chord $6x-24=0$, that is $x=4$. But the first circle has radius $4$ and centre the origin, so $x=4$ touches it at a single point: the circles touch rather than cross, and the common chord has length zero.

Checking with the distance rule: the second circle has centre $(3,0)$ and radius $1$, so $d=3=4-1$, internal contact. This is why the position of the circles should be settled before any chord formula is applied.

Illustration 9

Find the circle through the intersections of $x^2+y^2-6x+2=0$ and $x^2+y^2-4y-4=0$ that passes through $(1,1)$.

Use the family $S_1+\lambda S_2=0$. Substituting $(1,1)$ gives $S_1=1+1-6+2=-2$ and $S_2=1+1-4-4=-6$, so $-2-6\lambda =0$ and $\lambda =-\frac{1}{3}$. Multiplying out,

$3S_1-S_2=0\Rightarrow 2x^2+2y^2-18x+4y+10=0,$

that is $x^2+y^2-9x+2y+5=0$. The family device again avoids computing the intersections themselves.

Illustration 10

Find the radical centre of $x^2+y^2=9$, $x^2+y^2-4x=0$ and $x^2+y^2-6y+5=0$.

Taking the circles in pairs, the radical axes are $4x-9=0$, $6y-14=0$ and their combination. Solving the first two, the radical centre is $\left(\frac{9}{4},\frac{7}{3}\right)$, and the third axis passes through it automatically — which is the theorem, and also the check.

From this point the tangent lengths to all three circles are equal, so it is the centre of the unique circle cutting all three orthogonally.

8. Loci

A locus question asks for the equation satisfied by a moving point, and the method is always the same: call the point $(h,k)$, write every given condition in terms of $h$ and $k$, eliminate whatever parameter remains, and finally replace $h$ and $k$ by $x$ and $y$.

Illustration 11

Find the locus of the midpoints of chords of $x^2+y^2=a^2$ that subtend a right angle at the centre.

Such a chord has its endpoints at the ends of two perpendicular radii, so its length is $a\sqrt{2}$ and its distance from the centre is $\frac{a}{\sqrt{2}}$. The midpoint therefore lies at that fixed distance from the origin, giving

$x^2+y^2=\frac{a^2}{2}.$

Recognising that the condition fixes the distance from the centre replaces an algebraic elimination entirely.

Illustration 12

A variable line through the fixed point $(p,q)$ meets the axes at $A$ and $B$. Find the locus of the midpoint of $AB$.

If the midpoint is $(h,k)$, then $A=(2h,0)$ and $B=(0,2k)$, so the line is $\frac{x}{2h}+\frac{y}{2k}=1$. Passing through $(p,q)$ requires

$\frac{p}{2h}+\frac{q}{2k}=1\Rightarrow \frac{p}{x}+\frac{q}{y}=2,$

a rectangular hyperbola. Writing the intercepts in terms of the midpoint, rather than the other way round, is what keeps the elimination short.

Summary

An algebraic combination of two curve equations is a curve with a geometric meaning. Subtracting two circles gives the radical axis, the locus of equal tangent lengths, which exists whether or not the circles meet and coincides with the common chord when they do. Adding two lines as $L_1+\lambda L_2$ gives the family through their intersection, and adding two circles as $S_1+\lambda S_2$ gives the family through their common points, so neither intersection ever needs to be computed.

The parametric form $x=x_1+r\cos \theta$, $y=y_1+r\sin \theta$ turns any question about distances along a line into a question about the roots of a quadratic in $r$, with the sign of the product recording whether the fixed point lies inside the curve.

Both angle bisectors have the same equation up to a sign, so the sign must be chosen deliberately: normalise both constants to be positive and take the plus sign for the bisector containing the origin, then use the sign of $a_1{a}_2+b_1{b}_2$ to identify the acute one.

The four triangle centres each come from a different family of lines, and the orthocentre, centroid and circumcentre are always collinear with $H=3G-2O$, so any two of them give the third at once.

For circles, $S_1$ is the square of the tangent length and $T=0$ is the tangent at a point. Tangency conditions written as a quadratic in the slope let any angle-between-tangents question be answered from the relations between its roots.

The entire configuration of two circles — position and number of common tangents — follows from comparing the distance between the centres with the sum and difference of the radii, and that comparison should be made before any chord formula is used.