Lengths of Tangents from an External Point, Class 10

The equal tan­gents the­o­rem#

Set­ting up the pic­ture#

Let $O$ be the cen­tre of a cir­cle and $P$ a point out­side it. The two tan­gents from $P$ touch the cir­cle at $A$ and $B$. The seg­ments $PA$ and $PB$ are the lengths of the tan­gents from $P$. Join $OA$, $OB$ and $OP$. By the tan­gent–radius the­o­rem, $\angle OAP = \angle OBP = 90^\circ$.

State­ment#

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

Proof#

Given: A cir­cle with cen­tre $O$; $PA$ and $PB$ are tan­gents from an exter­nal point $P$, touch­ing the cir­cle at $A$ and $B$.

To prove: $PA = PB$.

Con­struc­tion: Join $OA$, $OB$ and $OP$.

Proof: In $\triangle OAP$ and $\triangle OBP$:

1. $\angle OAP = \angle OBP = 90^\circ$ (radius is per­pen­dic­u­lar to the tan­gent at the point of con­tact).
2. $OP = OP$ (com­mon hypotenuse).
3. $OA = OB$ (radii of the same cir­cle).

So $\triangle OAP \cong \triangle OBP$ by the RHS (right angle–hypotenuse–side) con­gru­ence rule. Hence, by CPCT (cor­re­spond­ing parts of con­gru­ent tri­an­gles), $PA = PB$. $\blacksquare$

Another way to see it. By Pythago­ras' the­o­rem in the two right tri­an­gles, $PA^2 = OP^2 - OA^2$ and $PB^2 = OP^2 - OB^2$. Since $OA = OB$, the right-hand sides are equal, so $PA^2 = PB^2$ and $PA = PB$.

Con­se­quences of the the­o­rem#

The line to the cen­tre bisects the angles#

From the same con­gru­ence, $\angle OPA = \angle OPB$ and $\angle AOP = \angle BOP$ (CPCT). So:

• $OP$ bisects the angle between the two tan­gents, $\angle APB$.
• $OP$ bisects the angle $\angle AOB$ sub­tended at the cen­tre by the two points of con­tact.
• The cen­tre there­fore lies on the bisec­tor of the angle between the two tan­gents.

The angle between the tan­gents#

In quadri­lat­eral $OAPB$, the four angles add up to $360^\circ$. Two of them are right angles, so

$$\angle APB + \angle AOB = 360^\circ - 90^\circ - 90^\circ = 180^\circ.$$

In words: the angle between the two tan­gents and the angle sub­tended by the points of con­tact at the cen­tre are sup­ple­men­tary.

The chord of con­tact#

Since $PA = PB$, tri­an­gle $PAB$ is isosce­les, so $\angle PAB = \angle PBA$. Also $OA = OB$ and $PA = PB$, so both $O$ and $P$ are equidis­tant from $A$ and $B$. There­fore $OP$ is the per­pen­dic­u­lar bisec­tor of the chord $AB$.

Method: solv­ing tan­gent prob­lems#

1. Draw the fig­ure and mark every point of con­tact.
2. Join the cen­tre to each point of con­tact and mark the right angles.
3. From each exter­nal point (or ver­tex), write the two equal tan­gent lengths with the same let­ter, for exam­ple $x$.
4. For angles, use the right angles, the angle sum of the quadri­lat­eral $OAPB$, and the isosce­les tri­an­gles $PAB$ and $OAB$.
5. For lengths, write each side as a sum of two tan­gent lengths and form equa­tions; or use Pythago­ras in a right tri­an­gle con­tain­ing a radius.
6. Solve, and check that every length is pos­i­tive and the pieces add up to the given sides.

Worked exam­ples#

Exam­ple 1: Angles made by two tan­gents#

$PA$ and $PB$ are tan­gents from $P$ to a cir­cle with cen­tre $O$, and $\angle APB = 70^\circ$. Find $\angle AOB$, $\angle OPA$, $\angle PAB$ and $\angle OAB$.

Solu­tion.

$$\displaystyle \begin{aligned} \angle AOB &= 180^\circ - \angle APB = 180^\circ - 70^\circ = 110^\circ \\ \angle OPA &= \tfrac{1}{2}\angle APB = 35^\circ \quad (OP \text{ bisects } \angle APB) \end{aligned}$$

In $\triangle PAB$, $PA = PB$, so $\angle PAB = \angle PBA$ and

$$\displaystyle \angle PAB = \frac{180^\circ - 70^\circ}{2} = 55^\circ.$$

Since $\angle OAP = 90^\circ$, we get $\angle OAB = 90^\circ - 55^\circ = 35^\circ$. (Check: $\triangle OAB$ is isosce­les with $\angle AOB = 110^\circ$, so each base angle is $\displaystyle \frac{180^\circ - 110^\circ}{2} = 35^\circ$.)

Answer: $\angle AOB = 110^\circ$, $\angle OPA = 35^\circ$, $\angle PAB = 55^\circ$, $\angle OAB = 35^\circ$.

Notice the pat­tern: $\angle OAB$ is always half of $\angle APB$.

Exam­ple 2: Chord of the larger of two con­cen­tric cir­cles#

Two con­cen­tric cir­cles have radii $5$ cm and $13$ cm. Find the length of the chord of the larger cir­cle that touches the smaller cir­cle.

Solu­tion. Let $O$ be the com­mon cen­tre and let the chord $AB$ of the larger cir­cle touch the smaller cir­cle at $P$.

1. $AB$ is a tan­gent to the smaller cir­cle at $P$, so $OP \perp AB$ and $OP = 5$ cm.
2. The per­pen­dic­u­lar from the cen­tre to a chord bisects the chord, so $AP = PB$.
3. In right tri­an­gle $OPB$, $OB = 13$ cm (radius of the larger cir­cle).

$$\begin{aligned} PB &= \sqrt{OB^2 - OP^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ cm} \\ AB &= 2 \times PB = 24 \text{ cm} \end{aligned}$$

Answer: The chord is $24$ cm long.

Exam­ple 3: Tan­gents inclined at 60 degrees#

Two tan­gents $PA$ and $PB$ to a cir­cle of radius $4$ cm are inclined to each other at $60^\circ$. Find $OP$ and the length of each tan­gent.

Solu­tion. $OP$ bisects $\angle APB$, so $\angle OPA = 30^\circ$. In right tri­an­gle $OAP$ (right-angled at $A$), $OA = 4$ cm is oppo­site the $30^\circ$ angle.

$$\displaystyle \begin{aligned} \sin 30^\circ &= \frac{OA}{OP} \;\Rightarrow\; \frac{1}{2} = \frac{4}{OP} \;\Rightarrow\; OP = 8 \text{ cm} \\ PA &= \sqrt{OP^2 - OA^2} = \sqrt{64 - 16} = \sqrt{48} = 4\sqrt{3} \text{ cm} \end{aligned}$$

Answer: $OP = 8$ cm and $PA = PB = 4\sqrt{3}$ cm. Here $\triangle PAB$ is equi­lat­eral, since it is isosce­les with a $60^\circ$ apex angle, so the chord $AB$ is also $4\sqrt{3}$ cm.

Exam­ple 4: A quadri­lat­eral cir­cum­scrib­ing a cir­cle#

A quadri­lat­eral $ABCD$ is drawn to cir­cum­scribe a cir­cle, that is, all four sides touch the cir­cle. Prove that $AB + CD = AD + BC$. Hence find $AD$ if $AB = 7$ cm, $BC = 9$ cm and $CD = 8$ cm.

Proof. Let the sides $AB$, $BC$, $CD$ and $DA$ touch the cir­cle at $P$, $Q$, $R$ and $S$ respec­tively. Tan­gents from the same exter­nal point are equal, so

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

Adding these four equa­tions,

$$\begin{aligned} AP + BP + CR + DR &= AS + BQ + CQ + DS \\ (AP + BP) + (CR + DR) &= (BQ + CQ) + (AS + DS) \\ AB + CD &= BC + AD. \end{aligned}$$

This proves the result. $\blacksquare$

Cal­cu­la­tion. $AD = AB + CD - BC = 7 + 8 - 9 = 6$ cm.

Answer: $AD = 6$ cm.

A con­se­quence. If a par­al­lel­o­gram $ABCD$ cir­cum­scribes a cir­cle, then $AB = CD$ and $BC = AD$, so the result gives $2AB = 2BC$, that is $AB = BC$. All four sides are equal, so the par­al­lel­o­gram is a rhom­bus.

Exam­ple 5: Tri­an­gle with an inscribed cir­cle#

A cir­cle is inscribed in tri­an­gle $ABC$ with $AB = 10$ cm, $BC = 12$ cm and $CA = 8$ cm. It touches $BC$, $CA$ and $AB$ at $D$, $E$ and $F$. Find $AF$, $BD$ and $CE$.

Solu­tion. Tan­gents from each ver­tex are equal. Let

$$AF = AE = x, \qquad BF = BD = y, \qquad CD = CE = z.$$

Writ­ing each side as a sum of two tan­gent lengths:

$$\begin{aligned} x + y &= AB = 10 \\ y + z &= BC = 12 \\ z + x &= CA = 8 \end{aligned}$$

Adding all three: $2(x + y + z) = 30$, so $x + y + z = 15$. Then

$$x = 15 - 12 = 3, \qquad y = 15 - 8 = 7, \qquad z = 15 - 10 = 5.$$

Check: $x + y = 10$, $y + z = 12$, $z + x = 8$. All three sides agree.

Answer: $AF = 3$ cm, $BD = 7$ cm, $CE = 5$ cm.

Exam­ple 6: Radius of the incir­cle of a right tri­an­gle#

Tri­an­gle $ABC$ is right-angled at $B$ with $AB = 6$ cm and $BC = 8$ cm. Find the radius of the cir­cle inscribed in it.

Solu­tion. By Pythago­ras, $AC = \sqrt{36 + 64} = 10$ cm. Let the incir­cle have cen­tre $O$ and radius $r$, and let it touch $AB$, $BC$ and $CA$ at $F$, $D$ and $E$.

1. $OF \perp AB$ and $OD \perp BC$ (radius $\perp$ tan­gent), and $\angle B = 90^\circ$. So $OFBD$ has three right angles and is a rec­tan­gle. Its adja­cent sides $OF$ and $OD$ are both equal to $r$, so it is a square, and $BF = BD = r$.
2. Tan­gents from $A$: $AE = AF = 6 - r$.
3. Tan­gents from $C$: $CE = CD = 8 - r$.
4. Since $E$ lies on $AC$: $AE + CE = AC$.

$$\begin{aligned} (6 - r) + (8 - r) &= 10 \\ 14 - 2r &= 10 \\ r &= 2 \text{ cm} \end{aligned}$$

Answer: The inra­dius is $2$ cm. In gen­eral, for a right tri­an­gle with legs $a$, $b$ and hypotenuse $c$, the same steps give $\displaystyle r = \frac{a + b - c}{2}$.

Com­mon mis­takes#

• Writ­ing "SAS" or "SSS" in the equal-tan­gents proof. The cor­rect rule is RHS, because the equal right angles are not between the known sides.
• For­get­ting to state why $\angle OAP = 90^\circ$. Write the rea­son "radius $\perp$ tan­gent at the point of con­tact" every time.
• Assum­ing $\angle APB = \angle AOB$ or that they add to $90^\circ$. They are sup­ple­men­tary: they add up to $180^\circ$.
• In tri­an­gle prob­lems, labelling the two tan­gent pieces from one ver­tex with dif­fer­ent let­ters. Pieces from the same ver­tex are equal.
• Using $AB + BC = CD + DA$ for a cir­cum­scribed quadri­lat­eral. The cor­rect pair­ing is oppo­site sides: $AB + CD = AD + BC$.
• In con­cen­tric cir­cle prob­lems, for­get­ting to dou­ble the half-chord at the end.

Try these#

1. Two con­cen­tric cir­cles have radii $6$ cm and $10$ cm. Find the length of the chord of the larger cir­cle that touches the smaller cir­cle. Answer: $16$ cm.
2. A quadri­lat­eral $ABCD$ cir­cum­scribes a cir­cle, with $AB = 11$ cm, $BC = 14$ cm and $CD = 9$ cm. Find $AD$. Answer: $6$ cm.
3. Tan­gents $PA$ and $PB$ from $P$ touch a cir­cle with cen­tre $O$, and $\angle AOB = 125^\circ$. Find $\angle APB$. Answer: $55^\circ$.
4. The incir­cle of tri­an­gle $ABC$ touches $BC$, $CA$, $AB$ at $D$, $E$, $F$. If $AB = 7$ cm, $BC = 9$ cm and $CA = 10$ cm, find $AF$, $BD$ and $CE$. Answer: $AF = 4$ cm, $BD = 3$ cm, $CE = 6$ cm.
5. Find the inra­dius of a right tri­an­gle whose legs are $5$ cm and $12$ cm. Answer: $2$ cm.
6. Two tan­gents to a cir­cle of radius $5$ cm are per­pen­dic­u­lar to each other. Find the length of each tan­gent and the dis­tance of their meet­ing point from the cen­tre. Answer: $5$ cm each; $5\sqrt{2}$ cm.

Key terms#

Exter­nal point

A point out­side the cir­cle; exactly two tan­gents can be drawn from it.

Length of a tan­gent

The dis­tance from the exter­nal point to the point of con­tact.

RHS con­gru­ence

Two right tri­an­gles are con­gru­ent if their hypotenuses and one pair of other sides are equal.

CPCT

Cor­re­spond­ing parts of con­gru­ent tri­an­gles are equal.

Con­cen­tric cir­cles

Cir­cles in the same plane with the same cen­tre and dif­fer­ent radii.

Cir­cum­scribed quadri­lat­eral

A quadri­lat­eral all of whose sides touch a cir­cle drawn inside it.

Incir­cle

The cir­cle inside a tri­an­gle that touches all three sides.

Chord of con­tact

The chord join­ing the two points of con­tact of the tan­gents from an exter­nal point.

Com­mon ques­tions#

Why is the con­gru­ence rule RHS and not SAS?#

The known equal parts are a right angle, the com­mon hypotenuse $OP$ and the radii $OA = OB$. The right angle is not between the hypotenuse and the radius, so SAS does not apply; RHS is exactly the rule for this sit­u­a­tion.

Is the angle between the tan­gents always sup­ple­men­tary to the angle at the cen­tre?#

Yes. The quadri­lat­eral $OAPB$ has two right angles at $A$ and $B$, so the remain­ing two angles, at $P$ and at $O$, must add up to $180^\circ$.

Does every quadri­lat­eral have a cir­cle touch­ing all four sides?#

No. The result $AB + CD = AD + BC$ holds when­ever such a cir­cle exists, so a quadri­lat­eral whose oppo­site sides do not have equal sums can­not cir­cum­scribe a cir­cle. A non-square rec­tan­gle, for exam­ple, has no such cir­cle.

How do I find tan­gent lengths in a tri­an­gle quickly?#

Add the three sides and halve the total to get $x + y + z$. The tan­gent length from a ver­tex is this half-sum minus the side oppo­site that ver­tex.

Which of these proofs are most impor­tant for the exam?#

Learn the equal-tan­gents proof, the proof that $AB + CD = AD + BC$ for a cir­cum­scribed quadri­lat­eral, and the result that a par­al­lel­o­gram cir­cum­scrib­ing a cir­cle is a rhom­bus. These appear often as 3-mark ques­tions.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Kise­lev, A. P. Kise­lev's Geom­e­try, Book I: Planime­try (adapted by A. Given­tal). Sum­iz­dat.
3. Cox­eter, H. S. M. and Gre­itzer, S. L. Geom­e­try Revis­ited. Math­e­mat­i­cal Asso­ci­a­tion of Amer­ica.
4. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.