Chapter 10: Circles (Geometry)

There is one tangent at every point of the circle, and a circle has infinitely many points.

Infinitely many.

(i) one (ii) secant (iii) two (iv) point of contact

OP ⊥ PQ (Theorem 10.1), so in right △OPQ: $PQ=\sqrt{O{Q}^2-O{P}^2}=\sqrt{144-25}=\sqrt{119}$ cm.

(D) √119 cm

Construction: draw any circle with centre O and a line $l$. Draw the radius OP perpendicular to $l$; the line through P parallel to $l$ is a tangent (it is perpendicular to OP). Any line parallel to $l$ that is closer to O than P cuts the circle in two points: a secant.

See the figure: the tangent at P and a secant, both parallel to l.

The radius is perpendicular to the tangent: $r=\sqrt{25^2-24^2}=\sqrt{49}=7$ cm.

(A) 7 cm

In quadrilateral OPTQ, ∠OPT = ∠OQT = 90°, so $\angle PTQ=360°-90°-90°-110°=70°$.

(B) 70°

OP bisects ∠APB (△OAP ≅ △OBP), so ∠OPA = 40°. In right △OAP, $\angle POA=90°-40°=50°$.

(A) 50°

Let AB be a diameter with centre O, and let $l$ and $m$ be the tangents at A and B.

By Theorem 10.1, OA ⊥ $l$ and OB ⊥ $m$, so $\angle OAl=\angle OBm=90°$. AB is a transversal to $l$ and $m$, and the alternate angles it makes with them are both 90°, hence equal.

Both tangents are perpendicular to the diameter AB, so the alternate angles are equal (90°) and the tangents are parallel.

Let AB be the tangent at P, and suppose the perpendicular to AB at P does not pass through the centre O. The radius OP is also perpendicular to AB (Theorem 10.1).

Then there would be two different lines through P, both perpendicular to AB, which is impossible. So the perpendicular at P is the line PO, which passes through the centre.

Only one line through P is perpendicular to the tangent, and the radius OP is such a line, so the perpendicular passes through O.

$r=\sqrt{5^2-4^2}=3$ cm (the radius is perpendicular to the tangent).

3 cm

AB touches the smaller circle at P, so OP ⊥ AB and P is the mid-point of AB (the perpendicular from the centre bisects a chord).

In right △OPB: $PB=\sqrt{5^2-3^2}=4$ cm, so $AB=2\times 4=8$ cm.

8 cm

Let the circle touch AB, BC, CD and DA at P, Q, R and S. Tangents from an external point are equal (Theorem 10.2):

$AP=AS$, $BP=BQ$, $CR=CQ$, $DR=DS$.

Adding: $(AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)$, i.e. $AB+CD=AD+BC$.

Using equal tangents from A, B, C and D and adding gives AB + CD = AD + BC.

Let XY touch the circle at P and X′Y′ at Q. Join OC.

In △OPA and △OCA: OP = OC (radii), OA common, AP = AC (tangents from A). So △OPA ≅ △OCA (SSS), and ∠POA = ∠COA. Similarly ∠QOB = ∠COB.

POQ is a straight line (the diameter joining the points of contact of parallel tangents), so $\angle POA+\angle COA+\angle COB+\angle QOB=180°$, i.e. $2(\angle COA+\angle COB)=180°$.

So $\angle AOB=\angle COA+\angle COB=90°$.

OA and OB bisect ∠POC and ∠QOC, which together make 180°, so ∠AOB = 90°.

Let PA and PB be tangents from P to a circle with centre O. Then OA ⊥ PA and OB ⊥ PB, so ∠OAP = ∠OBP = 90°.

In quadrilateral OAPB, the angles add up to 360°: $\angle APB+\angle AOB=360°-90°-90°=180°$.

∠APB + ∠AOB = 180°, so the angles are supplementary.

Let parallelogram ABCD circumscribe a circle. By Q8, $AB+CD=AD+BC$.

In a parallelogram, $AB=CD$ and $AD=BC$, so $2AB=2BC$, i.e. $AB=BC$.

So all four sides are equal ($AB=BC=CD=DA$).

AB + CD = AD + BC with opposite sides equal gives AB = BC, so the parallelogram is a rhombus.

Let the circle touch AB at F and AC at E, and let $AF=AE=x$. Then $BF=BD=8$ and $CE=CD=6$, so $AB=x+8$, $AC=x+6$ and $BC=14$.

Area by Heron's formula, with $s=x+14$:

$$\sqrt{(x+14)\cdot x\cdot 6\cdot 8}=\sqrt{48x(x+14)}$$

Area by splitting at the incentre O: $\frac{1}{2}\times 4\times (AB+BC+CA)=2\times 2(x+14)=4(x+14)$.

Equating and squaring: $48x(x+14)=16(x+14{)}^2\Rightarrow 3x=x+14\Rightarrow x=7$.

AB = 15 cm and AC = 13 cm.

Let ABCD circumscribe a circle with centre O, touching AB, BC, CD, DA at P, Q, R, S. Join O to the vertices and to P, Q, R, S.

As in Q9, each line from O to a vertex bisects the angle at O between the two radii to the points of contact there. So the eight angles at O form four equal pairs:

$\angle 1=\angle 2$ (at A), $\angle 3=\angle 4$ (at B), $\angle 5=\angle 6$ (at C), $\angle 7=\angle 8$ (at D).

They add to 360°, so $2(\angle 2+\angle 3+\angle 6+\angle 7)=360°$, where $\angle 2+\angle 3=\angle AOB$ and $\angle 6+\angle 7=\angle COD$.

So $\angle AOB+\angle COD=180°$, and similarly $\angle BOC+\angle DOA=180°$.

The radii and the lines to the vertices make four pairs of equal angles at O summing to 360°, so ∠AOB + ∠COD = 180° (and ∠BOC + ∠AOD = 180°).