Circles — chapter notes

Let $d$ be the distance of a line from the centre $O$ of a circle of radius $r$.

• $d > r$: the line has no common point with the circle.
• $d = r$: exactly one common point — the line is a tangent, and the common point is the point of contact.
• $d < r$: two common points — the line is a secant.

A tangent is the limiting position of a secant $PQ$ as $Q$ moves along the circle to $P$. At each point of a circle there is exactly one tangent.

Number of tangents from a point: none from a point inside, exactly one from a point on the circle, exactly two from a point outside.

Theorem 1. The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Proof. Let $XY$ be the tangent at $P$. Take any point $Q \ne P$ on $XY$. $Q$ lies outside the circle (otherwise $XY$ would be a secant), so $OQ$ meets the circle at $R$ with $OQ > OR = OP$. Hence $OP$ is the shortest segment from $O$ to $XY$, and the shortest segment from a point to a line is perpendicular to it: $OP \perp XY$.

Consequences.

• The perpendicular to a tangent at its point of contact passes through the centre.
• The tangents at the ends of a diameter are parallel; the distance between two parallel tangents is the diameter.

At the point of contact $P$ of a tangent from an outside point $Q$, $\triangle OPQ$ is right-angled at $P$, with hypotenuse $OQ$:

Example: radius $8$ m, $OQ = 17$ m gives the tangent $\sqrt{289 - 64} = 15$ m.

Angles. If $PQ$ is a chord and $PT$ the tangent at $P$, then $\angle OPQ = 90^\circ - \angle TPQ$, and the isosceles triangle $OPQ$ gives $\angle POQ = 2\angle TPQ$.

Example: $\angle POQ = 110^\circ$ gives $\angle TPQ = 55^\circ$.

Concentric circles. A chord of the larger circle that touches the smaller one is bisected at the point of contact: half-chord $= \sqrt{R^2 - r^2}$.

Theorem 2. The lengths of the tangents drawn from an external point to a circle are equal.

Proof. Let $PA$, $PB$ be the tangents from $P$; join $OA$, $OB$, $OP$. $\angle OAP = \angle OBP = 90^\circ$, $OA = OB$, and $OP$ is common, so $\triangle OAP \cong \triangle OBP$ (RHS) and $PA = PB$.

Consequences (from the same congruence):

• $OP$ bisects $\angle APB$ and $\angle AOB$;
• $\angle APB + \angle AOB = 180^\circ$ (angle sum of $OAPB$);
• $\triangle PAB$ is isosceles, $AB \perp OP$, and $\angle OAB = \dfrac12 \angle APB$;
• if $\angle APB = 90^\circ$, $OAPB$ is a square and $OP = r\sqrt2$; if $\angle APB = 60^\circ$, $OP = 2r$ and $\triangle PAB$ is equilateral.

Mark the two equal tangents from every vertex, then add.

Quadrilateral $ABCD$ touching the circle at $P, Q, R, S$: $AB + CD = BC + DA$, and the perimeter is $2(AB + CD)$.

Triangle $ABC$ with incircle touching $BC$, $CA$, $AB$ at $D$, $E$, $F$: if the tangent lengths are $x$ (from $A$), $y$ (from $B$), $z$ (from $C$), then $x + y = AB$, $y + z = BC$, $z + x = CA$, and $x + y + z = s$ (the semi-perimeter). So

Example: $AB = 10$, $BD = 6$, $DC = 8$ gives $AF = 4$ and $AC = 12$.

Area and inradius: joining the centre to the vertices splits the triangle into three triangles of height $r$, so area $= r \times s$.

• A parallelogram that circumscribes a circle is a rhombus ($2AB = 2BC$). So a rectangle round a circle must be a square.
• In a triangle right-angled at $C$ with legs $a$, $b$ and hypotenuse $c$, the inradius is $r = \dfrac{a + b - c}{2}$ (the corner at $C$ forms a square of side $r$). Example: $8$, $15$, $17$ gives $r = 3$.
• In an isosceles triangle round a circle, the circle touches the base at its midpoint, and the vertex, the centre and that midpoint are collinear.
• The two tangents at the ends of a chord $PQ$ meet on the perpendicular bisector of $PQ$, which passes through the centre.