A Bicycle Wheel Touches the Road at One Point, and That Point Is Special

A tangent is a straight line that touches a circle at exactly one point, called the point of contact, without cutting into it. A line that cuts the circle at two points is a secant.

From a point outside a circle you can draw two tangents; from a point on the circle, one; from a point inside, none.

This part covers the tangent-radius angle, equal tangents, touching circles and the alternate segment theorem.

The radius drawn to the point of contact is the shortest distance from the centre to the tangent line, so it meets the tangent at a right angle.

Why. Every other point of the tangent lies outside the circle, so it is farther from the centre than the radius. The shortest distance from a point to a line is the perpendicular, so $OA\perp$ tangent.

Worked example 1 — length. A circle has radius $5$ cm and a point $P$ is $13$ cm from the centre. The tangent from $P$ touches at $A$.

$$PA=\sqrt{O{P}^2-O{A}^2}=\sqrt{169-25}=\sqrt{144}=12\text{ cm}$$

Worked example 2 — angle. The tangent at $A$ meets $P$ with $\angle AOP=60^{\circ }$.

$$\angle OAP=90^{\circ }\angle APO=180^{\circ }-90^{\circ }-60^{\circ }=30^{\circ }$$

An everyday example. A bicycle wheel standing on a flat road touches it at one point, and the spoke to that point stands straight up, at right angles to the road.

The substance. The line perpendicular to a tangent at its point of contact always passes through the centre — a quick way to locate a centre.

**If $PA$ and $PB$ are tangents from an external point $P$, then $PA=PB$, and $OP$ bisects both $\angle APB$ and $\angle AOB$.

Why.** Triangles $OAP$ and $OBP$ have right angles at $A$ and $B$, a common hypotenuse $OP$ and $OA=OB$, so they are congruent by RHS.

Useful result. In the quadrilateral $OAPB$, two angles are $90^{\circ }$, so

$$\angle APB+\angle AOB=180^{\circ }$$

Worked example 1. $\angle APB=50^{\circ }$.

$$\angle AOB=130^{\circ }\angle OPA=25^{\circ }\angle PAB=\frac{180^{\circ }-50^{\circ }}{2}=65^{\circ }\angle OAB=90^{\circ }-65^{\circ }=25^{\circ }$$

Worked example 2 — circumscribed quadrilateral. A circle touches all four sides of $ABCD$, with $AB=6$, $BC=7$ and $CD=8$ cm. Equal tangents from each vertex give $AB+CD=BC+AD$:

$$6+8=7+ADAD=7\text{ cm}$$

An everyday example. Two straight roads from a village that just touch the edge of a circular lake reach their touching points after equal distances.

The substance. **Figure $OAPB$ is a kite**, with $OP$ as its line of symmetry.

When two circles touch, the point of contact lies on the straight line joining their centres; the distance between the centres is the sum of the radii if they touch externally and the difference if they touch internally.

- Touching externally — $d=r_1+r_2$
- Touching internally — $d=r_1-r_2$
- At the point of contact, both circles share a common tangent, perpendicular to the line of centres

Worked example 1. Circles of radii $5$ cm and $3$ cm touch. The distance between centres is $8$ cm if they touch externally and $2$ cm if internally.

Worked example 2. Three circles touch one another externally. The distances between their centres are $10$, $12$ and $14$ cm.

$$r_1+r_2=10r_2+r_3=12r_1+r_3=14$$

Adding: $2(r_1+r_2+r_3)=36$, so $r_1+r_2+r_3=18$.

$$r_3=18-10=8r_1=18-12=6r_2=18-14=4\text{ cm}$$

An everyday example. Coins laid touching each other on a table have their centres and the touching point in a straight line.

The link. Both radii to the contact point are perpendicular to the same tangent, which is why they line up.

The angle between a tangent and a chord drawn from the point of contact equals the angle that chord subtends in the alternate segment of the circle.

If $TA$ is a tangent at $A$, $AB$ is a chord and $C$ is a point on the arc on the other side of $AB$, then $\angle TAB=\angle ACB$.

Worked example 1. $\angle TAB=65^{\circ }$, so $\angle ACB=65^{\circ }$.

Worked example 2. In $△ABC$ inscribed in a circle, the tangent at $A$ makes $\angle TAB=70^{\circ }$ and $\angle ABC=50^{\circ }$.

$$\angle ACB=70^{\circ }\angle BAC=180^{\circ }-70^{\circ }-50^{\circ }=60^{\circ }$$

Worked example 3 — with the centre. $\angle AOB=110^{\circ }$.

$$\angle ACB=55^{\circ }\angle OAB=\frac{180^{\circ }-110^{\circ }}{2}=35^{\circ }\angle TAB=90^{\circ }-35^{\circ }=55^{\circ }$$

Both routes agree.

An everyday example. A straight road leaving a circular roundabout at a tangent makes an angle with any straight path across the roundabout that matches the angle seen from the far side.

The substance. The angle is always in the segment on the opposite side of the chord from the angle between tangent and chord.