Tangent to a Circle and the Radius Theorem, Class 10

A line and a cir­cle: three pos­si­bil­i­ties#

Take a cir­cle and a straight line drawn in the same plane. Slide the line across the page and watch how many points it shares with the cir­cle. Only three things can hap­pen.

Non-inter­sect­ing line#

The line has no com­mon point with the cir­cle. It passes com­pletely out­side. Its dis­tance from the cen­tre is greater than the radius.

Secant#

The line meets the cir­cle at two dis­tinct points, say $A$ and $B$. Such a line is called a secant. The part $AB$ that lies inside the cir­cle is a chord. The dis­tance of a secant from the cen­tre is less than the radius.

Tan­gent#

The line meets the cir­cle at exactly one point, say $P$. Such a line is called a tan­gent to the cir­cle, and $P$ is called the point of con­tact. The word comes from the Latin word for "touch­ing". The dis­tance of a tan­gent from the cen­tre is exactly equal to the radius.

From secant to tan­gent#

Imag­ine a secant through $A$ and $B$ that you slowly move par­al­lel to itself, away from the cen­tre. The two points $A$ and $B$ come closer and closer. At one posi­tion they merge into a sin­gle point. At that moment the secant has become a tan­gent. So a tan­gent can be thought of as the spe­cial case of a secant whose two points of inter­sec­tion coin­cide.

One tan­gent at each point#

At any given point on a cir­cle there is one and only one tan­gent. You can check this with a thread and a coin: how­ever you try, only one straight line touches the coin at a cho­sen point on its rim with­out cut­ting across it. The the­o­rem below explains why: the tan­gent at $P$ must be the line through $P$ per­pen­dic­u­lar to $OP$, and through a given point there is only one line per­pen­dic­u­lar to a given line.

The tan­gent–radius the­o­rem#

State­ment#

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

In sym­bols: if $XY$ is the tan­gent at $P$ to a cir­cle with cen­tre $O$, then $OP \perp XY$, that is, $\angle OPX = \angle OPY = 90^\circ$.

Proof#

Given: A cir­cle with cen­tre $O$ and a tan­gent $XY$ touch­ing it at $P$.

To prove: $OP \perp XY$.

Con­struc­tion: Take any point $Q$ on $XY$ other than $P$, and join $OQ$.

Proof:

1. A tan­gent meets the cir­cle at only one point, namely $P$. So $Q$, which is on the tan­gent but is not $P$, can­not lie on the cir­cle.
2. $Q$ can­not lie inside the cir­cle either. If it did, the line $XY$ would pass from inside the cir­cle to out­side it and would cut the cir­cle at two points, so it would be a secant and not a tan­gent. Hence $Q$ lies out­side the cir­cle.
3. For a point out­side the cir­cle, its dis­tance from the cen­tre is greater than the radius. So $OQ \gt OP$.
4. This holds for every point $Q$ of $XY$ other than $P$. There­fore $OP$ is the short­est of all the seg­ments join­ing $O$ to points of the line $XY$.
5. Of all the seg­ments that can be drawn from a point to a line, the short­est is the per­pen­dic­u­lar one. Hence $OP \perp XY$.

This com­pletes the proof. $\blacksquare$

Use­ful con­se­quences#

• The line drawn through the point of con­tact per­pen­dic­u­lar to the tan­gent passes through the cen­tre. (This line is some­times called the nor­mal at that point.)
• The per­pen­dic­u­lar drawn at the end of a radius, at the point where it meets the cir­cle, is a tan­gent.
• Tan­gents drawn at the two ends of a diam­e­ter are par­al­lel. (See Exam­ple 5.)
• When­ever a radius meets a tan­gent at its point of con­tact, you get a right angle, so a right-angled tri­an­gle is usu­ally wait­ing to be used.

How many tan­gents can you draw from a point?#

The answer depends on where the point lies.

When the point $P$ is out­side the cir­cle and the tan­gent from $P$ touches the cir­cle at $Q$, the seg­ment $PQ$ is called the length of the tan­gent from $P$.

Find­ing the length of a tan­gent#

The key for­mula#

Let $O$ be the cen­tre, $r$ the radius, and $P$ an exter­nal point with $OP = d$. If the tan­gent from $P$ touches the cir­cle at $Q$, then $\angle OQP = 90^\circ$ by the the­o­rem, so tri­an­gle $OQP$ is right-angled at $Q$ with hypotenuse $OP$. By Pythago­ras' the­o­rem,

$$PQ^2 = OP^2 - OQ^2, \qquad \text{so} \qquad PQ = \sqrt{d^2 - r^2}.$$

The same tri­an­gle also gives $r = \sqrt{d^2 - PQ^2}$ and $d = \sqrt{r^2 + PQ^2}$. Because $PQ$ is a real length, $d \gt r$, which agrees with the fact that $P$ is out­side the cir­cle.

Method in steps#

1. Draw a neat fig­ure. Mark the cen­tre $O$, the exter­nal point $P$ and the point of con­tact $Q$.
2. Join $OQ$ and mark the right angle at $Q$. Write "radius $\perp$ tan­gent" as your rea­son.
3. Iden­tify the hypotenuse. It is always $OP$, the side oppo­site the right angle.
4. Write Pythago­ras' the­o­rem for tri­an­gle $OQP$ and sub­sti­tute the two known lengths.
5. Solve for the unknown, take the pos­i­tive square root, and write the answer with units.
6. If an angle is given instead of a length, use the trigono­met­ric ratios in the same right tri­an­gle.

Worked exam­ples#

Exam­ple 1: Length of the tan­gent#

A point $P$ is $13$ cm from the cen­tre $O$ of a cir­cle of radius $5$ cm. Find the length of the tan­gent drawn from $P$ to the cir­cle.

Solu­tion. Let the tan­gent touch the cir­cle at $Q$. Then $OQ = 5$ cm and $OP = 13$ cm. Since the radius is per­pen­dic­u­lar to the tan­gent at the point of con­tact, $\angle OQP = 90^\circ$.

$$\begin{aligned} PQ^2 &= OP^2 - OQ^2 \\ &= 13^2 - 5^2 \\ &= 169 - 25 = 144 \\ PQ &= 12 \text{ cm} \end{aligned}$$

Answer: The length of the tan­gent is $12$ cm.

Exam­ple 2: Find­ing the radius#

From a point $Q$, the length of the tan­gent to a cir­cle is $24$ cm, and the dis­tance of $Q$ from the cen­tre is $25$ cm. Find the radius of the cir­cle.

Solu­tion. Let the tan­gent touch the cir­cle at $T$ and let the radius be $r$. Then $\angle OTQ = 90^\circ$, $QT = 24$ cm and $OQ = 25$ cm.

$$\begin{aligned} OT^2 &= OQ^2 - QT^2 \\ &= 625 - 576 = 49 \\ r = OT &= 7 \text{ cm} \end{aligned}$$

Answer: The radius is $7$ cm.

Exam­ple 3: When an angle is given#

$PQ$ is a tan­gent at $P$ to a cir­cle with cen­tre $O$ and radius $6$ cm. The point $Q$ on the tan­gent is such that $\angle OQP = 30^\circ$. Find $OQ$ and $PQ$.

Solu­tion. Since $OP$ is a radius and $PQ$ is the tan­gent at $P$, $\angle OPQ = 90^\circ$. In right tri­an­gle $OPQ$, the side $OP = 6$ cm is oppo­site the $30^\circ$ angle.

$$\displaystyle \begin{aligned} \sin 30^\circ &= \frac{OP}{OQ} \;\Rightarrow\; \frac{1}{2} = \frac{6}{OQ} \;\Rightarrow\; OQ = 12 \text{ cm} \\ \tan 30^\circ &= \frac{OP}{PQ} \;\Rightarrow\; \frac{1}{\sqrt{3}} = \frac{6}{PQ} \;\Rightarrow\; PQ = 6\sqrt{3} \text{ cm} \end{aligned}$$

Check with Pythago­ras: $PQ^2 = 12^2 - 6^2 = 144 - 36 = 108$, and $(6\sqrt{3})^2 = 36 \times 3 = 108$. The val­ues agree.

Answer: $OQ = 12$ cm and $PQ = 6\sqrt{3} \approx 10.39$ cm.

Exam­ple 4: Near­est dis­tance from a point to the cir­cle#

The tan­gent from a point $A$ to a cir­cle of radius $15$ cm has length $20$ cm. Find (a) the dis­tance of $A$ from the cen­tre $O$, and (b) the short­est dis­tance from $A$ to the cir­cle.

Solu­tion. (a) Let the tan­gent touch the cir­cle at $T$. Then $\angle OTA = 90^\circ$ and

$$OA = \sqrt{OT^2 + AT^2} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25 \text{ cm}.$$

(b) The point of the cir­cle near­est to $A$ lies on the seg­ment $OA$, at dis­tance $15$ cm from $O$. So the short­est dis­tance from $A$ to the cir­cle is

$$OA - r = 25 - 15 = 10 \text{ cm}.$$

Answer: (a) $25$ cm; (b) $10$ cm.

Exam­ple 5: Tan­gents at the ends of a diam­e­ter#

Prove that the tan­gents drawn at the two ends of a diam­e­ter of a cir­cle are par­al­lel.

Solu­tion. Let $AB$ be a diam­e­ter of a cir­cle with cen­tre $O$. Let $l$ be the tan­gent at $A$ and $m$ the tan­gent at $B$.

1. $OA$ is a radius and $l$ is the tan­gent at $A$, so $OA \perp l$. Since $O$ lies on $AB$, this means $AB \perp l$.
2. Sim­i­larly $OB \perp m$, so $AB \perp m$.
3. Now the lines $l$ and $m$ are both per­pen­dic­u­lar to the same line $AB$. The trans­ver­sal $AB$ makes an angle of $90^\circ$ with each, so the two inte­rior angles on the same side add up to $90^\circ + 90^\circ = 180^\circ$.
4. When the inte­rior angles on the same side of a trans­ver­sal are sup­ple­men­tary, the lines are par­al­lel. Hence $l \parallel m$.

Answer: The tan­gents at the ends of a diam­e­ter are par­al­lel. $\blacksquare$

A use­ful fol­low-up: the dis­tance between these two par­al­lel tan­gents equals the diam­e­ter $2r$.

Com­mon mis­takes#

• Tak­ing the tan­gent $PQ$ as the hypotenuse. The right angle is at the point of con­tact, so the hypotenuse is always $OP$, the line from the cen­tre to the exter­nal point.
• Adding the squares when you should sub­tract. To find the tan­gent or the radius, sub­tract: $PQ^2 = OP^2 - r^2$. Add only when you are find­ing $OP$.
• Mark­ing the right angle at the exter­nal point $P$ or at the cen­tre $O$ instead of at the point of con­tact.
• Say­ing that one tan­gent can be drawn from a point inside the cir­cle. The cor­rect num­ber is zero.
• Writ­ing the short­est dis­tance from an exter­nal point to the cir­cle as $OP$. It is $OP - r$.
• Quot­ing the the­o­rem with­out the rea­son in a proof. Always write "radius is per­pen­dic­u­lar to the tan­gent at the point of con­tact" beside the right angle.

Try these#

1. A point is $15$ cm from the cen­tre of a cir­cle of radius $9$ cm. Find the length of the tan­gent from the point. Answer: $12$ cm.
2. The tan­gent from a point $10$ cm from the cen­tre of a cir­cle has length $8$ cm. Find the radius. Answer: $6$ cm.
3. The radius of a cir­cle is $7$ cm and the tan­gent from a point $P$ is $24$ cm long. How far is $P$ from the cen­tre? Answer: $25$ cm.
4. $PQ$ is the tan­gent at $P$ to a cir­cle of radius $5$ cm and $\angle OQP = 45^\circ$. Find $PQ$ and $OQ$. Answer: $PQ = 5$ cm, $OQ = 5\sqrt{2}$ cm.
5. From a point $P$, the tan­gent to a cir­cle of radius $12$ cm is $35$ cm long. Find the short­est dis­tance from $P$ to the cir­cle. Answer: $25$ cm (since $OP = 37$ cm).
6. How many tan­gents can be drawn to a cir­cle from a point on it, and how many from a point inside it? Answer: One and zero.

Key terms#

Secant

A line that meets a cir­cle at two dis­tinct points.

Tan­gent

A line that meets a cir­cle at exactly one point.

Non-inter­sect­ing line

A line that has no point in com­mon with the cir­cle.

Point of con­tact

The sin­gle com­mon point of a cir­cle and its tan­gent.

Length of a tan­gent

The length of the seg­ment from an exter­nal point to the point of con­tact.

Exter­nal point

A point whose dis­tance from the cen­tre is greater than the radius.

Nor­mal

The line through the point of con­tact per­pen­dic­u­lar to the tan­gent; it passes through the cen­tre.

Com­mon ques­tions#

Why is the tan­gent per­pen­dic­u­lar to the radius?#

Every point of the tan­gent other than the point of con­tact lies out­side the cir­cle, so it is far­ther from the cen­tre than the radius. The radius to the point of con­tact is there­fore the short­est seg­ment from the cen­tre to the tan­gent, and the short­est seg­ment from a point to a line is the per­pen­dic­u­lar.

Can a tan­gent pass through a point inside the cir­cle?#

No. Any line through an inside point cuts the cir­cle twice, so it is a secant. That is why no tan­gent can be drawn from an inside point.

Is the length of a tan­gent the same as the dis­tance from the point to the cir­cle?#

No. The length of the tan­gent is $\sqrt{d^2 - r^2}$, while the short­est dis­tance from the point to the cir­cle is $d - r$. For $d = 13$ and $r = 5$ these are $12$ and $8$.

Which the­o­rems from this topic should I be able to prove in the exam?#

The two stan­dard proofs are the tan­gent–radius the­o­rem proved here and the the­o­rem that the two tan­gents from an exter­nal point are equal in length. Both are fre­quently asked as 3-mark ques­tions.

Do I need a fig­ure in every tan­gent ques­tion?#

It is strongly rec­om­mended. A clear fig­ure with the right angle marked at the point of con­tact shows the exam­iner your rea­son­ing and helps you pick the cor­rect hypotenuse.

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. Heath, T. L. (trans.) The Thir­teen Books of Euclid's Ele­ments. Dover Pub­li­ca­tions.
4. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.