Circles: Tangents, Secants and Two Circles

Tangent to a Circle

A tangent is a line that touches a circle at exactly one point, called the point of contact. A secant cuts the circle at two points. You can picture a tangent as a secant whose two cutting points have slid together into one.

How many tangents from a point?

• From a point inside the circle: none, because every line through it cuts the circle twice.
• From a point on the circle: exactly one.
• From a point outside the circle: exactly two.

The tangent is perpendicular to the radius at the point of contact. If $T$ is the point of contact, then $OT$ is at 90° to the tangent. Every other point of the tangent lies outside the circle, so it is farther from $O$ than $T$ is. The shortest segment from $O$ to a line is the perpendicular, so $OT$ must be it. This works backwards too: a line drawn at right angles to a radius, through its outer end, is a tangent.

Tangent length. From an outside point $P$ at distance $d = OP$ from the centre, the tangent $PT$ is a side of the right triangle $OTP$, with the right angle at $T$ and $OP$ as the hypotenuse:

$$PT = \sqrt{OP^2 - r^2} = \sqrt{d^2 - r^2}$$

• $OP = 10$ cm and $r = 6$ cm: $PT = \sqrt{100 - 36} = 8$ cm.
• $PT = 24$ cm and $r = 7$ cm: $OP = \sqrt{576 + 49} = 25$ cm.
• $PT = 8$ cm and $OP = 10$ cm: $r = \sqrt{100 - 64} = 6$ cm.

Parallel tangents. The tangents at the two ends of a diameter are both perpendicular to that diameter, so they are parallel, and the gap between them is the diameter $2r$. So a circle that fits between two parallel lines 12 cm apart, touching both, has radius 6 cm.

[SSC Tip] The right angle sits at the point of contact, not at the centre, so in $PT = \sqrt{d^2 - r^2}$ the distance $OP$ is the hypotenuse. A wrong option usually adds the two squares instead of subtracting.

Two Tangents from an Outside Point

From a point $P$ outside the circle you can draw two tangents, touching the circle at $A$ and $B$.

The two tangents are equal: $PA = PB$. Triangles $OAP$ and $OBP$ are right-angled at $A$ and $B$, share the hypotenuse $OP$, and have $OA = OB$, so they are congruent (RHS). The same congruence gives two more facts:

• $OP$ bisects $\angle APB$, and it bisects $\angle AOB$ too.
• $OP$ is the perpendicular bisector of the chord $AB$, which is called the chord of contact.

Angle at $P$ + angle at $O$ = 180°. In the quadrilateral $OAPB$, the angles at $A$ and $B$ are 90° each, so

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

If the tangents meet at 50°, the radii to the points of contact meet at 130°. Since the angles at $A$ and $B$ add to 180°, $OAPB$ is a cyclic quadrilateral, with $OP$ as a diameter of its circle.

Two angles worth remembering.

• $\angle APB = 60^\circ$: then $\angle OPA = 30^\circ$, so in the 30°-60°-90° triangle $OAP$, $OP = 2r$ and $PA = r\sqrt{3}$. With $r = 5$ cm: $OP = 10$ cm and $PA = 5\sqrt{3}$ cm.
• $\angle APB = 90^\circ$: then $OAPB$ has four right angles and $PA = OA$, so it is a square, with $PA = r$ and $OP = r\sqrt{2}$. With $r = 4$ cm, $PA = 4$ cm.

[Tier 2] Length of the chord of contact. Half of $AB$ is the altitude from $A$ to the hypotenuse in the right triangle $OAP$, so

$$AB = \frac{2 \times PA \times r}{OP}$$

With $OP = 10$ and $r = 6$: $PA = 8$ and $AB = \frac{2 \times 8 \times 6}{10} = 9.6$.

"Tangents from an outside point are equal" settles a surprising number of questions by itself, especially perimeter questions. Mark the equal pieces with ticks on your rough figure as soon as you see two tangents from one point.

Angles Made by Tangents

Take two tangents $PA$ and $PB$ with $\angle APB = \theta$. Every other angle in the picture follows from $\theta$.

• $\angle AOB = 180^\circ - \theta$.
• Triangle $PAB$ is isosceles ($PA = PB$), so $\angle PAB = \angle PBA = 90^\circ - \frac{\theta}{2}$.
• $\angle OAB = 90^\circ - \angle PAB = \frac{\theta}{2}$. So the angle between a radius and the chord of contact is half the angle between the tangents.
• For $C$ on the major arc: $\angle ACB = \frac{1}{2}\angle AOB = 90^\circ - \frac{\theta}{2}$, the same as $\angle PAB$.
• For $D$ on the minor arc: $\angle ADB = 180^\circ - \angle ACB = 90^\circ + \frac{\theta}{2}$.

Take $\theta = 40^\circ$: $\angle AOB = 140^\circ$, $\angle PAB = 70^\circ$, $\angle OAB = 20^\circ$, $\angle ACB = 70^\circ$ and $\angle ADB = 110^\circ$. Going backwards is just as quick. If $\angle ACB = 65^\circ$ for $C$ on the major arc, then $\angle AOB = 130^\circ$ and the tangents meet at $50^\circ$.

$\angle PAB = \angle ACB$ is no accident: it is the alternate segment theorem, which has its own part below.

[Tier 2] Through the arcs. Like two chords meeting outside the circle, the angle between two tangents is half the difference of the two arcs between the points of contact: $\theta = \frac{1}{2}\left(\text{major arc} - \text{minor arc}\right)$. With $\theta = 40^\circ$ the arcs are 220° and 140°, and $\frac{220^\circ - 140^\circ}{2} = 40^\circ$.

[SSC Tip] Three numbers keep coming back: $\theta$, $180^\circ - \theta$ and $90^\circ - \frac{\theta}{2}$. In most tangent-angle questions the answer is one of them and the other two are usually among the options, so name the angle you want before you calculate.

Concentric Circles and Parallel Tangents

A chord of the bigger circle that touches the smaller one. Take two circles with the same centre $O$, with radii $R$ (bigger) and $r$. Let a chord $AB$ of the bigger circle touch the smaller circle at $M$. Then $OM \perp AB$ (radius and tangent), and a perpendicular from the centre bisects a chord, so $M$ is the midpoint of $AB$ and

$$AB = 2\sqrt{R^2 - r^2}$$

• $R = 10$ cm and $r = 6$ cm: $AB = 2\sqrt{100 - 36} = 16$ cm.
• A chord of 72 cm in a circle of radius 39 cm touches a smaller circle with the same centre. Half the chord is 36, so $r = \sqrt{1521 - 1296} = 15$ cm.

Every such chord has the same length, because each one is exactly $r$ from the centre. The area of the ring between the two circles is a Chapter 13 question, and this half-chord $\sqrt{R^2 - r^2}$ is the piece it needs.

Two parallel tangents and a third. The tangents at the ends of a diameter $AB$ are parallel. Let a third tangent, touching the circle at $C$, cut them at $X$ and $Y$.

Then $XA = XC$ and $YB = YC$ (tangents from one point), so $OX$ bisects $\angle AOC$ and $OY$ bisects $\angle BOC$. Those two angles together make the straight angle $AOB$, so

$$\angle XOY = \frac{1}{2} \times 180^\circ = 90^\circ$$

wherever $C$ is. Also $XY = XA + YB$, and since $OC$ is the altitude of the right triangle $XOY$, $XC \times CY = r^2$.

"Tangents at the ends of a diameter" means "parallel tangents, $2r$ apart". And a tangent cutting across two parallel tangents always makes a right angle at the centre; SSC asks this as a one-line question.

Tangent and Secant from One Point

From an outside point $P$, draw a tangent $PT$ and a secant that cuts the circle at $A$ (nearer) and $B$ (farther). Then

$$PT^2 = PA \times PB$$

Why: triangles $PTA$ and $PBT$ share the angle at $P$, and $\angle PTA = \angle PBT$ by the alternate segment theorem (further down this section). So the triangles are similar, $\frac{PT}{PB} = \frac{PA}{PT}$, and cross-multiplying gives the rule.

• $PA = 5$ cm and $AB = 15$ cm: $PB = 20$ cm, so $PT = \sqrt{5 \times 20} = 10$ cm.
• $PT = 15$ cm and $PA = 9$ cm: $PB = \frac{225}{9} = 25$ cm, so the chord $AB = 16$ cm.

Two secants from one point. If two secants from $P$ cut the circle at $A$, $B$ and at $C$, $D$ (with $A$ and $C$ the nearer points), both products equal $PT^2$, so

$$PA \times PB = PC \times PD$$

With $PA = 3$, $PB = 10$ and $PC = 2$: $PD = 15$, so the chord $CD = 13$.

This is the same rule as for chords crossing inside the circle (in Circles: Chords, Arcs and Angles), now with $P$ outside. For an outside point, every such product equals $OP^2 - r^2$, which is $PT^2$ by Pythagoras.

[SSC Tip] Multiply whole lengths measured from $P$: $PA \times PB$, never $PA \times AB$. If the question gives the chord $AB$, add $PA$ to it first. Using $AB$ in place of $PB$ is the standard wrong option.

The Alternate Segment Theorem

The angle between a tangent and a chord, at the point of contact, equals the angle that the chord makes at any point of the circle on the other side of the chord (the alternate segment).

In the figure, the tangent $XY$ at $A$ and the chord $AB$ make $\angle BAY = x$. The alternate segment is the part of the circle on the far side of $AB$ from $Y$, so for $C$ there, $\angle ACB = x$.

Why: the radius $OA$ is at 90° to the tangent, so $\angle OAB = 90^\circ - x$. Triangle $OAB$ is isosceles, so $\angle AOB = 180^\circ - 2(90^\circ - x) = 2x$, and a point on the major arc sees half of that, which is $x$.

So the angle between a tangent and a chord is also half the angle the chord makes at the centre. A point on the near arc (on the same side as the angle $x$) sees the chord at $180^\circ - x$.

• The tangent at $A$ makes 55° with chord $AB$. Then the angle in the alternate segment is 55°, $\angle AOB = 110^\circ$, and a point on the other arc sees $AB$ at 125°.
• Triangle $ABC$ sits in a circle, and the tangent at $A$ makes 50° with $AB$ on one side and 60° with $AC$ on the other. Then $\angle ACB = 50^\circ$, $\angle ABC = 60^\circ$ and $\angle BAC = 70^\circ$. The three angles at $A$ along the tangent add to 180°, as they should.

With a diameter. Let $AB$ be a diameter, and let the tangent at a point $C$ meet $AB$ produced beyond $B$ at $P$. By this theorem $\angle PCB = \angle CAB$, and $\angle ABC = 90^\circ - \angle CAB$. Since $\angle ABC$ is the exterior angle of triangle $PBC$,

$$\angle P = 90^\circ - 2\angle CAB$$

For $\angle CAB = 25^\circ$, $\angle P = 40^\circ$.

Match the angle to the correct segment: the angle on one side of the chord at the tangent equals the angle in the segment on the other side. Pick the wrong side and you get $180^\circ - x$, which is always among the options.

Quadrilateral Circumscribing a Circle

When all four sides of $ABCD$ touch a circle, the quadrilateral circumscribes the circle, and the circle is inscribed in it. Each corner sends two equal tangents to the circle.

Call the tangent lengths from $A$, $B$, $C$, $D$ by $a$, $b$, $c$, $d$. Then $AB = a + b$, $BC = b + c$, $CD = c + d$ and $DA = d + a$, so

$$AB + CD = AD + BC = a + b + c + d$$

The two pairs of opposite sides have equal sums, and the perimeter is $2(AB + CD)$.

• $AB = 6$ cm, $BC = 10$ cm and $CD = 13$ cm: $AD = 6 + 13 - 10 = 9$ cm, and the perimeter is $2 \times 19 = 38$ cm.

Shapes that can hold a circle. A parallelogram circumscribing a circle has $AB = CD$ and $BC = AD$, so the rule gives $2AB = 2BC$ and all four sides are equal: it must be a rhombus. In the same way a rectangle circumscribing a circle is a square. In a trapezium circumscribing a circle, both parallel sides are tangents, so the height equals the diameter.

[Tier 2] Angles at the centre. The lines from the centre to the four corners bisect the four angles, and opposite sides make supplementary angles at the centre: $\angle AOB + \angle COD = 180^\circ$. (Each of these angles is $180^\circ$ minus half of two corner angles, and the four corner angles add to 360°.)

You rarely need the separate tangent lengths; the opposite-sides rule does the work in one line. Watch the wording: a circle "inscribed in a quadrilateral" uses this rule, while a quadrilateral "inscribed in a circle" is cyclic and uses opposite angles adding to 180°.

A Circle Inside a Triangle: Tangent Lengths

The incircle of triangle $ABC$ touches $BC$, $CA$ and $AB$ at $D$, $E$ and $F$. Equal tangents from each corner split every side into two tangent lengths.

With $a = BC$, $b = CA$, $c = AB$ and the semi-perimeter $s = \frac{a + b + c}{2}$:

$$AF = AE = s - a \qquad BD = BF = s - b \qquad CD = CE = s - c$$

Each corner's tangent length is $s$ minus the side opposite that corner. For $BC = 14$, $CA = 15$ and $AB = 13$: $s = 21$, so $AF = 7$, $BD = 6$ and $CD = 8$. Check: $BD + CD = 14 = BC$.

Right triangle. At the right angle $C$, the two tangents and two radii form a square, so the tangents from $C$ are each $r$. The tangents from $A$ and $B$ are then $b - r$ and $a - r$, and together they make the hypotenuse: $(a - r) + (b - r) = c$. So

$$r = \frac{a + b - c}{2}$$

Legs 8 and 15 with hypotenuse 17: $r = \frac{8 + 15 - 17}{2} = 3$. The general formula $r = \frac{\Delta}{s}$ is in Centres of a Triangle.

The third tangent. Tangents $PA$ and $PB$ touch a circle. A third tangent, at a point $R$ on the minor arc, cuts $PA$ at $X$ and $PB$ at $Y$. Then $XR = XA$ and $YR = YB$, so

$$PX + XY + YP = PX + XA + YB + YP = PA + PB = 2PA$$

The perimeter of triangle $PXY$ is twice the tangent length, wherever $R$ sits. With $PA = 9$ cm, the perimeter is 18 cm. ([Tier 2] The circle here is the excircle of triangle $PXY$, so $PA$ equals the semi-perimeter of that triangle.)

"$s$ minus the opposite side" is easy to mix up: the tangent from $A$ uses $a = BC$, the side that does not touch $A$. Check by adding two tangent lengths to get a side.

Two Circles: Touching, Cutting and Common Tangents

Take two circles with radii $R \ge r$ and centres $d$ apart. Comparing $d$ with $R + r$ and $R - r$ tells you how they sit, and how many lines touch both of them (their common tangents).

Circles with the same centre ($d = 0$) fall in the last row.

When two circles touch, the point of contact lies on the line joining the centres, and the common tangent at that point is perpendicular to this line.

Take $R = 7$ cm and $r = 3$ cm. If $d = 10$ cm they touch externally (3 common tangents); $d = 4$ cm, they touch internally (1); $d = 6$ cm, they cut each other (2); $d = 12$ cm, they are apart (4); and $d = 2$ cm, the small circle sits inside without touching (0).

Direct and transverse. A direct common tangent has both circles on the same side of it. A transverse one passes between them and crosses the line of centres. Transverse tangents need the circles to be apart (then there are two) or touching externally (then they merge into the single tangent at the point of contact).

The touching cases get asked both ways. "Two circles of radii 9 cm and 5 cm touch internally; find the distance between their centres" is 9 − 5 = 4 cm, not 14 cm. Internal means subtract.

Direct and Transverse Common Tangents

For circles with radii $R$ and $r$ whose centres are $d$ apart, the length of a common tangent is measured between its two points of contact:

$$\text{direct: } L = \sqrt{d^2 - (R - r)^2} \qquad \text{transverse: } L = \sqrt{d^2 - (R + r)^2}$$

Why the direct formula: the radii $O_1A$ and $O_2B$ are both perpendicular to the tangent $AB$. Draw a line from $O_2$ parallel to $AB$, meeting $O_1A$ at $N$. Then $ABO_2N$ is a rectangle, so $O_2N = AB$ and $O_1N = R - r$. Pythagoras in triangle $O_1NO_2$ gives $AB^2 = d^2 - (R - r)^2$.

For a transverse tangent the two radii point opposite ways, so the same construction meets $O_1A$ extended, and $O_1N = R + r$.

Take $R = 6$ cm, $r = 2$ cm and $d = 10$ cm: the direct tangent is $\sqrt{100 - 16} = \sqrt{84} = 2\sqrt{21}$ cm and the transverse one is $\sqrt{100 - 64} = 6$ cm. The direct tangent is always the longer of the two.

Where the tangents cross the line of centres. A transverse tangent cuts $O_1O_2$ at a point $Q$ that divides it in the ratio $R : r$. Here $O_1Q = 10 \times \frac{6}{8} = 7.5$ cm and $O_2Q = 2.5$ cm. The two direct tangents meet on $O_1O_2$ extended beyond the smaller circle, at a point $X$ with $\frac{XO_1}{XO_2} = \frac{R}{r} = 3$. Since $XO_1 - XO_2 = 10$, $XO_2 = 5$ cm. Both facts come from similar right triangles (radius to tangent at each circle).

[SSC Tip] Direct goes with $R - r$, transverse with $R + r$. The standard trap swaps them. A quick check: the transverse tangent is the shorter one, and it exists only when $d > R + r$ (or is zero when the circles touch).

Touching Circles and Three Circles

Two circles touching externally. When $d = R + r$, the direct common tangent becomes

$$AB = \sqrt{(R + r)^2 - (R - r)^2} = \sqrt{4Rr} = 2\sqrt{Rr}$$

For radii 8 cm and 2 cm, $AB = 2\sqrt{16} = 8$ cm.

Now let the circles touch at $P$, and let the common tangent at $P$ meet $AB$ at $M$. From $M$ there are two tangents to each circle: $MA = MP$ and $MP = MB$. So $M$ is the midpoint of $AB$, $MP = \sqrt{Rr}$, and $P$ lies on the circle with diameter $AB$. That gives

$$\angle APB = 90^\circ$$

for any two circles touching externally. In the same way $MO_1$ and $MO_2$ bisect the two angles at $M$, which add to 180°, so $\angle O_1MO_2 = 90^\circ$ as well.

Three circles touching each other externally. The centres form a triangle whose sides are the sums of pairs of radii. So the perimeter of that triangle is $2(r_1 + r_2 + r_3)$. Going back, each radius is the semi-perimeter minus the side opposite its centre, just like tangent lengths in a triangle.

Let the centres be $A$, $B$, $C$ with $AB = 5$, $BC = 6$ and $CA = 7$. Then $s = 9$, and the radii are $s - BC = 3$ at $A$, $s - CA = 2$ at $B$ and $s - AB = 4$ at $C$. Check: $3 + 2 = 5$, $2 + 4 = 6$ and $3 + 4 = 7$. Three equal circles of radius $r$ touching one another have centres forming an equilateral triangle of side $2r$.

[Tier 2] A small circle in the gap. Two circles of radii $r_1$ and $r_2$ touch each other and a straight line. A smaller circle in the gap, touching both of them and the same line, has radius $r_3$ with

$$\frac{1}{\sqrt{r_3}} = \frac{1}{\sqrt{r_1}} + \frac{1}{\sqrt{r_2}}$$

This follows from the touching-circle tangent $2\sqrt{Rr}$ measured three times along the line: $2\sqrt{r_1 r_3} + 2\sqrt{r_3 r_2} = 2\sqrt{r_1 r_2}$. For two circles of radius 4: $\frac{1}{\sqrt{r_3}} = \frac{1}{2} + \frac{1}{2} = 1$, so $r_3 = 1$.

For touching circles, $2\sqrt{Rr}$ is faster than the full formula. And "$\angle APB = 90^\circ$" is a favourite one-mark question whose options include 60° and "cannot be found".

Circles in Squares, Triangles and Other Circles

These "fit one shape inside another" questions only need the distance from the centre to a side or to a corner.

Square of side $a$. The circle inside touches all four sides, so its diameter is $a$ and $r = \frac{a}{2}$. The circle outside passes through the corners, so its diameter is the diagonal $a\sqrt{2}$ and $R = \frac{a}{\sqrt{2}}$. So $R : r = \sqrt{2} : 1$. For a square of side 10 cm, $r = 5$ cm and $R = 5\sqrt{2}$ cm. Turned around, the biggest square that fits in a circle of radius $R$ has side $R\sqrt{2}$.

Triangles. In a right triangle, $r = \frac{a + b - c}{2}$ and $R = \frac{c}{2}$ (the hypotenuse is a diameter). In an equilateral triangle of side $a$, $r = \frac{a}{2\sqrt{3}}$ and $R = \frac{a}{\sqrt{3}} = 2r$. The other centres are in Centres of a Triangle.

Largest circle in a semicircle. In a semicircle of radius $R$, the largest circle touches the diameter at its midpoint and the arc at the top, so its diameter is $R$ and its radius is $\frac{R}{2}$.

[Tier 2] Largest circle in a quadrant. In a quarter circle of radius $R$, the circle touching both straight edges and the arc has its centre on the line that bisects the corner, at distance $r\sqrt{2}$ from the corner. Adding $r$ reaches the arc:

$$r\sqrt{2} + r = R \;\Rightarrow\; r = \frac{R}{\sqrt{2} + 1} = R(\sqrt{2} - 1)$$

For $R = 10$ cm, $r = 10(\sqrt{2} - 1)$, about 4.14 cm.

Circles inside a circle. Two equal circles inside a circle of radius $R$, touching each other and the big circle, each have radius $\frac{R}{2}$: their two diameters fill one diameter of the big circle. [Tier 2] For three equal circles touching one another and the big circle, the centres form an equilateral triangle of side $2r$, whose centre is $\frac{2r}{\sqrt{3}}$ from each corner. Adding $r$ gives $R$, so $r = \frac{\sqrt{3}R}{2 + \sqrt{3}} = (2\sqrt{3} - 3)R$.

Whenever a circle sits in a right-angled corner touching both sides, its centre is $r\sqrt{2}$ from the corner. That single fact solves the quadrant and most other "circle in a corner" questions.

Solved Examples

Question 1

A point $P$ is 61 cm from the centre of a circle of radius 11 cm. What is the length of the tangent drawn from $P$ to the circle? (a) 50 cm (b) $\sqrt{3842}$ cm (c) 60 cm (d) 72 cm

Answer. The radius meets the tangent at a right angle, so $OP$ is the hypotenuse of the right triangle. Tangent $= \sqrt{61^2 - 11^2} = \sqrt{3721 - 121} = \sqrt{3600}$ = (c) 60 cm.

Question 2

The tangent from a point $P$ to a circle is 20 cm long, and $P$ is 29 cm from the centre. What is the radius of the circle? (a) 21 cm (b) 9 cm (c) $\sqrt{1241}$ cm (d) 49 cm

Answer. The triangle made by the radius, the tangent and $OP$ is right-angled at the point of contact, with hypotenuse $OP = 29$. So $r = \sqrt{29^2 - 20^2} = \sqrt{841 - 400} = \sqrt{441}$ = (a) 21 cm.

Question 3

$PA$ and $PB$ are tangents from $P$ to a circle with centre $O$. If $\angle APB = 64^\circ$, what is $\angle OAB$? (a) 58° (b) 116° (c) 64° (d) 32°

Answer. The angle at $P$ and the angle at $O$ add up to 180°, so $\angle AOB = 116^\circ$. Triangle $OAB$ has $OA = OB$, so $\angle OAB = \frac{180^\circ - 116^\circ}{2}$ = (d) 32°. That is half of $\angle APB$, as expected. 58° is $\angle PAB$, the angle between the tangent and the chord.

Question 4

Two tangents drawn from a point $P$ to a circle of radius 9 cm make an angle of 60° with each other. What is the length of each tangent? (a) 18 cm (b) $9\sqrt{3}$ cm (c) 9 cm (d) $3\sqrt{3}$ cm

Answer. $OP$ cuts the 60° in half, so $\angle OPA = 30^\circ$, and triangle $OAP$ has its right angle at $A$. In a 30°-60°-90° triangle the side opposite 30° is half the hypotenuse: $OA = 9$, so $OP = 18$ cm, and the side opposite 60° is $\sqrt{3}$ times the short side: $PA$ = (b) $9\sqrt{3}$ cm.

Question 5

$PA$ and $PB$ are tangents to a circle with centre $O$, and $\angle APB = 72^\circ$. $C$ is a point on the major arc $AB$. What is $\angle ACB$? (a) 36° (b) 108° (c) 54° (d) 126°

Answer. First the centre: $\angle AOB = 180^\circ - 72^\circ = 108^\circ$. $C$ is on the major arc, so it sees half of that: $\angle ACB$ = (c) 54°. Quick form: $90^\circ - \frac{72^\circ}{2}$. 126° is what a point on the minor arc would see.

Question 6

Two circles with the same centre have radii 50 cm and 14 cm. What is the length of a chord of the larger circle that touches the smaller circle? (a) 96 cm (b) 48 cm (c) 72 cm (d) $4\sqrt{674}$ cm

Answer. The chord touches the small circle, so the small circle's radius meets it at right angles, and that perpendicular from the centre cuts the chord in half. Half the chord is $\sqrt{50^2 - 14^2} = \sqrt{2304} = 48$ cm, so the chord is (a) 96 cm.

Question 7

From a point $P$, a tangent $PT$ and a secant $PAB$ are drawn to a circle. If $PA = 4$ cm and $AB = 5$ cm, what is the length of $PT$? (a) $2\sqrt{5}$ cm (b) 3 cm (c) 4.5 cm (d) 6 cm

Answer. The whole secant is $PB = PA + AB = 9$ cm. Then $PT^2 = PA \times PB = 4 \times 9 = 36$, so $PT$ = (d) 6 cm. Using $PA \times AB = 20$ gives $2\sqrt{5}$ cm, the usual slip.

Question 8

Two secants $PAB$ and $PCD$ are drawn to a circle from an outside point $P$. If $PA = 5$ cm, $AB = 7$ cm and $PC = 6$ cm, what is the length of $CD$? (a) 10 cm (b) 4 cm (c) $\frac{35}{6}$ cm (d) 6 cm

Answer. Use whole lengths from $P$: $PB = 5 + 7 = 12$ cm. Then $PC \times PD = PA \times PB = 60$, so $PD = 10$ cm and $CD = PD - PC = 10 - 6$ = (b) 4 cm.

Question 9

A tangent $PT$ and a secant $PAB$ are drawn from a point $P$ to a circle. If $PT = 14$ cm and $PA = 7$ cm, what is the length of the chord $AB$? (a) 21 cm (b) 28 cm (c) 14 cm (d) 7 cm

Answer. $PT^2 = PA \times PB$, so $196 = 7 \times PB$ and $PB = 28$ cm. The chord is the part inside the circle: $AB = 28 - 7$ = (a) 21 cm.

Question 10

The tangent at a point $A$ of a circle makes an angle of 58° with the chord $AB$. What angle does $AB$ make at a point $D$ on the minor arc $AB$? (a) 58° (b) 116° (c) 122° (d) 32°

Answer. By the alternate segment theorem, a point on the major arc (the far side of the chord) sees $AB$ at 58°. $D$ is on the minor arc, the other side of the chord, so it sees $180^\circ - 58^\circ$ = (c) 122°.

Question 11

$AB$ is a diameter of a circle, and the tangent at a point $C$ on the circle meets $AB$ produced at $P$, beyond $B$. If $\angle CAB = 32^\circ$, what is $\angle CPA$? (a) 32° (b) 58° (c) 64° (d) 26°

Answer. By the alternate segment theorem, $\angle PCB = \angle CAB = 32^\circ$. $AB$ is a diameter, so $\angle ACB = 90^\circ$ and $\angle ABC = 58^\circ$. Now $\angle ABC$ is the exterior angle of triangle $PBC$ at $B$, so $58^\circ = \angle CPB + 32^\circ$, which gives $\angle CPA$ = (d) 26°. Shortcut: $90^\circ - 2 \times 32^\circ = 26^\circ$.

Question 12

Triangle $ABC$ is drawn in a circle. The tangent at $A$ makes an angle of 70° with $AB$ on one side and 48° with $AC$ on the other side. What is $\angle BAC$? (a) 70° (b) 62° (c) 48° (d) 118°

Answer. The three angles at $A$ along the straight tangent add up to 180°: $70^\circ + \angle BAC + 48^\circ = 180^\circ$, so $\angle BAC$ = (b) 62°. As a check, the alternate segment theorem gives $\angle ACB = 70^\circ$ and $\angle ABC = 48^\circ$, and $62^\circ + 70^\circ + 48^\circ = 180^\circ$.

Question 13

A quadrilateral $ABCD$ is drawn around a circle so that all four sides touch it. If $AB = 8$ cm, $BC = 11$ cm and $CD = 10$ cm, what is the length of $AD$? (a) 7 cm (b) 13 cm (c) 9 cm (d) 29 cm

Answer. For a quadrilateral around a circle, the opposite sides have equal sums: $AB + CD = AD + BC$. So $8 + 10 = AD + 11$, and $AD$ = (a) 7 cm.

Question 14

The incircle of triangle $ABC$ touches $BC$ at $D$. If $AB = 9$ cm, $BC = 11$ cm and $CA = 12$ cm, what is $BD$? (a) 7 cm (b) 5.5 cm (c) 4 cm (d) 5 cm

Answer. Half the perimeter is $s = \frac{9 + 11 + 12}{2} = 16$ cm. The tangent from $B$ is $s$ minus the side opposite $B$, which is $CA$: $BD = 16 - 12$ = (c) 4 cm. Check: $CD = 16 - 9 = 7$, and $4 + 7 = 11 = BC$.

Question 15

$PA$ and $PB$ are tangents from $P$ to a circle, each 12 cm long. A third tangent touches the circle at a point $R$ of the minor arc $AB$ and cuts $PA$ at $X$ and $PB$ at $Y$. What is the perimeter of triangle $PXY$? (a) 12 cm (b) 36 cm (c) 18 cm (d) 24 cm

Answer. Tangents from one point are equal, so $XA = XR$ and $YB = YR$. The perimeter of $PXY$ is $PX + XR + RY + YP = PX + XA + YB + YP = PA + PB$. Both tangents are 12 cm, so the perimeter is (d) 24 cm, wherever $R$ sits on the arc.

Question 16

A right-angled triangle has legs 28 cm and 45 cm. What is the radius of the circle that touches all three of its sides? (a) 20 cm (b) 10 cm (c) 26.5 cm (d) 7 cm

Answer. The hypotenuse is $\sqrt{28^2 + 45^2} = \sqrt{784 + 2025} = \sqrt{2809} = 53$ cm. For a right triangle, $r = \frac{a + b - c}{2} = \frac{28 + 45 - 53}{2}$ = (b) 10 cm.

Question 17

The hypotenuse of a right-angled triangle is 26 cm and the radius of its incircle is 4 cm. What is the perimeter of the triangle? (a) 60 cm (b) 34 cm (c) 52 cm (d) 68 cm

Answer. $r = \frac{a + b - c}{2}$ gives $a + b = 2r + c = 8 + 26 = 34$ cm. The perimeter is $a + b + c = 34 + 26$ = (a) 60 cm. (The legs are 10 cm and 24 cm, and $10^2 + 24^2 = 26^2$.)

Question 18

A circle of radius 7 cm is drawn inside a square, touching all four sides. Another circle passes through all four corners of the square. What is the radius of the second circle? (a) 14 cm (b) $7\sqrt{3}$ cm (c) $7\sqrt{2}$ cm (d) $14\sqrt{2}$ cm

Answer. The inner circle touches opposite sides, so the side of the square is its diameter, 14 cm. The outer circle passes through opposite corners, so its diameter is the diagonal, $14\sqrt{2}$ cm, and its radius is (c) $7\sqrt{2}$ cm. $14\sqrt{2}$ cm is the diameter, not the radius.

Question 19

Two circles of radii 27 cm and 12 cm touch each other externally. What is the length of their direct common tangent? (a) 39 cm (b) 18 cm (c) 15 cm (d) 36 cm

Answer. Touching externally means the centres are $27 + 12 = 39$ cm apart. The direct common tangent is $\sqrt{39^2 - (27 - 12)^2} = \sqrt{1521 - 225} = \sqrt{1296}$ = (d) 36 cm. Shortcut for touching circles: $2\sqrt{27 \times 12} = 2 \times 18 = 36$.

Question 20

Two circles of radii 7 cm and 2 cm have their centres 15 cm apart. What is the length of a transverse common tangent? (a) $10\sqrt{2}$ cm (b) 12 cm (c) 9 cm (d) $2\sqrt{14}$ cm

Answer. A transverse tangent uses the sum of the radii: $\sqrt{15^2 - (7 + 2)^2} = \sqrt{225 - 81} = \sqrt{144}$ = (b) 12 cm. $10\sqrt{2}$ cm is the direct common tangent, which uses $7 - 2$.

Question 21

Two circles touch each other internally. The radius of the bigger circle is 15 cm and the distance between the centres is 6 cm. What are the radius of the smaller circle and the number of common tangents? (a) 9 cm and 3 (b) 21 cm and 2 (c) 9 cm and 1 (d) 21 cm and 1

Answer. Touching internally means $d = R - r$, so $6 = 15 - r$ and $r = 9$ cm. Circles touching internally have just one common tangent, the line through the point of contact. So the answer is (c) 9 cm and 1.

Question 22

The direct common tangent of two circles of radii 11 cm and 4 cm is 24 cm long. How far apart are their centres? (a) 25 cm (b) $\sqrt{801}$ cm (c) 24 cm (d) 15 cm

Answer. $24^2 = d^2 - (11 - 4)^2$, so $d^2 = 576 + 49 = 625$ and $d$ = (a) 25 cm.

Question 23

Two circles touch each other externally at $P$. A direct common tangent touches them at $A$ and $B$. What is $\angle APB$? (a) 60° (b) 90° (c) 120° (d) It depends on the radii

Answer. Let the common tangent at $P$ meet $AB$ at $M$. From $M$, $MA$ and $MP$ are tangents to the first circle, so $MA = MP$; and $MP$ and $MB$ are tangents to the second, so $MP = MB$. So $M$ is the midpoint of $AB$ with $MP = \frac{AB}{2}$, which puts $P$ on the circle with diameter $AB$. The angle in a semicircle gives $\angle APB$ = (b) 90°, whatever the radii are.

Question 24

Three circles touch one another externally. The distances between their centres are 10 cm, 13 cm and 15 cm. What is the radius of the largest circle? (a) 6 cm (b) 7.5 cm (c) 19 cm (d) 9 cm

Answer. Call the radii $x$, $y$ and $z$. Touching externally means $x + y = 10$, $y + z = 13$ and $z + x = 15$. Adding all three, $2(x + y + z) = 38$, so $x + y + z = 19$. Take away each pair in turn: the radii are $19 - 13 = 6$, $19 - 15 = 4$ and $19 - 10 = 9$. The largest is (d) 9 cm.

Question 25

[Tier 2] From a point $P$ that is 50 cm from the centre of a circle of radius 40 cm, tangents $PA$ and $PB$ are drawn. What is the length of the chord $AB$? (a) 24 cm (b) 48 cm (c) 30 cm (d) 60 cm

Answer. The tangent length is $PA = \sqrt{2500 - 1600} = 30$ cm. $OP$ is the perpendicular bisector of $AB$, so half of $AB$ is the altitude from $A$ in the right triangle $OAP$: $\frac{OA \times PA}{OP} = \frac{40 \times 30}{50} = 24$ cm. So $AB$ = (b) 48 cm.