A Line Touching a Circle Once Must Meet the Radius at a Right Angle

Take a circle and a straight line cutting through it. The line meets the circle at two points. Now slide the line outward, away from the centre, and watch those two points travel toward each other. At one particular position they meet and become a single point, and one more nudge outward and the line misses the circle entirely.

That one special position is a tangent, and the whole chapter is about what makes it special.

- A secant cuts the circle at two points
- A tangent touches the circle at exactly one point, called the point of contact
- Any further out, the line and the circle have no common point at all

So a tangent is the limiting case of a secant, the moment when its two intersection points merge. That is worth holding on to, because it explains why a tangent behaves so differently from every other line near a circle.

Two results carry the entire chapter, and both are provable in a few lines:

- The tangent at any point of a circle is perpendicular to the radius through the point of contact
- The lengths of the two tangents drawn from an external point are equal

Together they turn almost every circle problem into a right-triangle problem, which means the Pythagoras theorem does the arithmetic. A tangent length, a radius and the distance from the centre always form a right triangle, and knowing two of them gives the third.

The applications are where the marks are: quadrilaterals drawn around a circle, the angle between two tangents, and chords of concentric circles. Each is one line of reasoning followed by one line of arithmetic.

This page covers the CBSE Class 10 Maths chapter on circles: secants and tangents, the two tangent theorems and their proofs, and applications to circumscribed figures and concentric circles.

It depends entirely on where the point is, and there are exactly three cases.

- A point inside the circle — no tangent can be drawn. Every line through an interior point must cut the circle twice, so every such line is a secant
- A point on the circle — exactly one tangent, namely the tangent at that very point
- A point outside the circle — exactly two tangents, one on each side

Why an interior point admits none. A tangent must have exactly one point in common with the circle. A line through a point inside must enter and leave, so it meets the circle twice and is a secant by definition. No amount of tilting changes that, which is why the answer is zero rather than one.

Why an exterior point admits exactly two. Rotate a line about the external point. Far from the circle it misses; as it turns it starts cutting the circle; turning further it misses again on the other side. It passes through the touching position exactly twice — once on each side — giving two tangents.

The vocabulary to get right, because questions use it precisely.

- Tangent — the line itself, which is infinite
- Point of contact — the single point where it touches
- Length of the tangent from an external point — the distance from that point to the point of contact, which is a segment, not the whole line

That last distinction matters. A tangent line has no length; the length of the tangent from a point means the segment between the point and the point of contact, and that is the quantity all the numericals ask for.

A related fact about a circle and a line, stated in terms of distance. If $d$ is the distance from the centre to the line and $r$ is the radius:

- **$d<r$ — the line is a secant, cutting at two points
- $d=r$ — the line is a tangent, touching at one point
- $d>r$ — the line misses the circle

The tangent case is exactly $d=r$, which is another way of saying that the perpendicular from the centre to a tangent is the radius — the theorem of the next section in disguise.

One boundary case worth naming. The two tangents from an external point are distinct unless the point is on the circle, where they coincide into one. And the two tangents drawn at the ends of a diameter are parallel**, since each is perpendicular to the same line. That is a standard two-mark proof: both tangents are perpendicular to the diameter, and two lines perpendicular to the same line are parallel.

Because the radius to the point of contact is the shortest distance from the centre to the tangent, and the shortest distance from a point to a line is always along the perpendicular.

The proof, which is short and worth writing out. Let a circle have centre $O$ and let a tangent touch it at $P$. Take any other point $Q$ on the tangent, distinct from $P$.

- **$Q$ lies outside the circle**, because the tangent meets the circle only at $P$
- **So $OQ$ is longer than the radius**: $OQ>OP$
- **This holds for every such $Q$**, so $OP$ is the shortest of all the distances from $O$ to points of the tangent
- The shortest distance from a point to a line is along the perpendicular

**Hence $OP$ is perpendicular to the tangent, which is what was to be proved.

Notice where the tangent's defining property entered. It was used exactly once, in the first step: because the line touches at only one point, every other point on it is outside the circle. A secant would fail there, since points of a secant between its two intersections lie inside.

The converse is also a theorem and is sometimes what a question wants. A line drawn through the end of a radius and perpendicular to it is a tangent to the circle. So "perpendicular to the radius at its end" and "tangent" are two descriptions of the same line.

And the practical consequence: a right triangle appears in every tangent figure.** If $P$ is the point of contact and $Q$ is an external point on the tangent, then $△OPQ$ is right-angled at $P$, so

$$O{Q}^2=O{P}^2+P{Q}^2$$

which rearranges to give the length of the tangent from $Q$:

$$PQ=\sqrt{O{Q}^2-r^2}$$

Worked example 1. A tangent touches a circle of radius $5$ cm at $P$, and meets a line through the centre $O$ at a point $Q$ with $OQ=13$ cm. Find $PQ$.

$$PQ=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12\text{ cm}$$

Check: $5$, $12$ and $13$ form a Pythagorean triple, so the triangle closes exactly. Correct.

Worked example 2 — finding the radius instead. 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.

$$r=\sqrt{25^2-24^2}=\sqrt{625-576}=\sqrt{49}=7\text{ cm}$$

Check: $7$, $24$, $25$ is another triple. Correct.

Worked example 3 — finding the distance from the centre. The length of a tangent from a point $A$ to a circle of radius $3$ cm is $4$ cm. How far is $A$ from the centre?

$$OA=\sqrt{3^2+4^2}=\sqrt{9+16}=5\text{ cm}$$

Notice which side is the hypotenuse in all three. The distance from the centre to the external point is always the hypotenuse, because the right angle is at the point of contact. Treating the tangent length as the hypotenuse is the error this section exists to prevent, and it always produces an answer that is too small.

Worked example 4 — a tangent at the end of a diameter. Prove that the tangents at the two ends of a diameter of a circle are parallel.

Let $AB$ be a diameter with tangents at $A$ and at $B$. By the theorem, the tangent at $A$ is perpendicular to $OA$ and the tangent at $B$ is perpendicular to $OB$. **But $OA$ and $OB$ lie along the same straight line $AB$. Two lines perpendicular to the same line are parallel, so the two tangents are parallel.**

Because the two right triangles they form share a hypotenuse and have equal radii, so they are congruent.

The proof. Let $P$ be an external point and let $PA$ and $PB$ be the two tangents to a circle with centre $O$, touching at $A$ and $B$. Join $OA$, $OB$ and $OP$.

In $△OAP$ and $△OBP$:

- $\angle OAP=\angle OBP=90^{\circ }$, since a tangent is perpendicular to the radius at the point of contact
- $OA=OB$, both being radii
- $OP=OP$, common to both triangles

So the two triangles are congruent by the RHS criterion, and therefore $PA=PB$.

Three more equalities come free from the same congruence, and questions ask for each of them:

- $\angle OPA=\angle OBP$'s partner $\angle OPB$ — so **$OP$ bisects the angle between the two tangents**
- $\angle AOP=\angle BOP$ — so **$OP$ bisects the angle between the two radii
- $OP$ is the perpendicular bisector of the chord $AB$, the chord of contact

Worked example 1 — the angle between two tangents.** Tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at $80^{\circ }$. Find $\angle POA$.

Since $OP$ bisects $\angle APB$,

$$\angle APO=\frac{80^{\circ }}{2}=40^{\circ }$$

In $△OAP$, the angle at $A$ is $90^{\circ }$, so

$$\angle POA=180^{\circ }-90^{\circ }-40^{\circ }=50^{\circ }$$

Check by the other route. In quadrilateral $OAPB$ the four angles are $\angle OAP=90^{\circ }$, $\angle OBP=90^{\circ }$, $\angle APB=80^{\circ }$ and $\angle AOB$, and they sum to $360^{\circ }$:

$$\angle AOB=360^{\circ }-90^{\circ }-90^{\circ }-80^{\circ }=100^{\circ }$$

and since $OP$ bisects it, $\angle POA=50^{\circ }$. The two routes agree.

And notice the general relation hidden in that check. $\angle APB+\angle AOB=180^{\circ }$, so the angle between two tangents and the angle between the two radii are supplementary. That one line answers a great many questions directly.

Worked example 2 — the distance from the centre. The angle between two tangents drawn from an external point $P$ to a circle of radius $a$ is $60^{\circ }$. Find $OP$.

$OP$ bisects the $60^{\circ }$ angle, so in the right triangle $OAP$ the angle at $P$ is $30^{\circ }$, the side $OA=a$ is opposite it, and $OP$ is the hypotenuse:

$$\sin 30^{\circ }=\frac{a}{OP}⟹\frac{1}{2}=\frac{a}{OP}⟹OP=2a$$

Check with the tangent length. Then $PA=\sqrt{4a^2-a^2}=a\sqrt{3}$, and $\tan 30^{\circ }$ should be $\frac{a}{a\sqrt{3}}=\frac{1}{\sqrt{3}}$. It is.

Worked example 3 — a proof about the chord of contact. Prove that $\angle APB=2\angle OAB$.

Since $PA=PB$, the triangle $APB$ is isosceles, so $\angle PAB=\angle PBA$. If $\angle APB=2\theta$ then

$$\angle PAB=\frac{180^{\circ }-2\theta }{2}=90^{\circ }-\theta$$

But $\angle OAP=90^{\circ }$, so

$$\angle OAB=90^{\circ }-\angle PAB=90^{\circ }-(90^{\circ }-\theta )=\theta$$

**Hence $\angle APB=2\theta =2\angle OAB$, as required.

The pattern in all of these. Every one began by drawing the radii to the two points of contact and the line to the centre, which splits the figure into two congruent right triangles. Draw those three lines first and the rest is angle chasing** — and a figure without them is very hard to reason about.

Label every tangent length from the same vertex with the same letter. Almost every problem then becomes a short piece of algebra.

Result 1 — a quadrilateral circumscribing a circle. If a circle touches all four sides of quadrilateral $ABCD$, then

$$AB+CD=AD+BC$$

Why. Let the circle touch $AB$, $BC$, $CD$ and $DA$ at $P$, $Q$, $R$ and $S$. The two tangents from each vertex are equal, so write $AP=AS=p$, $BP=BQ=q$, $CQ=CR=r$ and $DR=DS=s$. Then

$$AB+CD=(p+q)+(r+s)$$
$$AD+BC=(p+s)+(q+r)$$

**Both are $p+q+r+s$, so they are equal. The proof is nothing more than labelling and adding.

Worked example 1.** A circle touches all four sides of a quadrilateral $ABCD$. If $AB=6$ cm, $BC=7$ cm and $CD=4$ cm, find $AD$.

$$AB+CD=AD+BC⟹6+4=AD+7$$
$$AD=3\text{ cm}$$

Check: $6+4=10$ and $3+7=10$. The two pairs of opposite sides give the same total.

A useful corollary. If the quadrilateral is a parallelogram, then $AB=CD$ and $AD=BC$, so the relation $AB+CD=AD+BC$ becomes $AB+AB=AD+AD$, giving $AB=AD$. A parallelogram that circumscribes a circle must be a rhombus, which is a standard one-mark result.

Result 2 — the inradius of a right triangle. For a right triangle, the radius of the inscribed circle is

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

where $a$ and $b$ are the legs and $c$ the hypotenuse.

Worked example 2. A circle is inscribed in a right triangle $ABC$ right-angled at $B$, with $AB=6$ cm and $BC=8$ cm. Find its radius.

First the hypotenuse:

$$AC=\sqrt{36+64}=\sqrt{100}=10\text{ cm}$$
$$r=\frac{6+8-10}{2}=\frac{4}{2}=2\text{ cm}$$

Check with tangent lengths, which is also how the formula is derived. The circle touches $AB$ and $BC$, and since the angle at $B$ is a right angle and the two radii to those points of contact are perpendicular to them, the tangent lengths from $B$ are both $r=2$ cm. Then the tangent length from $A$ is $6-2=4$ cm and from $C$ is $8-2=6$ cm. **So the hypotenuse should be $4+6=10$ cm, and it is. Correct.

Result 3 — chords of concentric circles. A chord of the larger of two concentric circles that touches the smaller one is bisected at the point of contact.

Why. The chord is a tangent to the smaller circle, so the radius of the smaller circle to the point of contact is perpendicular to it. But a perpendicular from the centre to a chord bisects the chord — and the two circles share the same centre.

Worked example 3.** Two concentric circles have radii $5$ cm and $3$ cm. Find the length of the chord of the larger circle which touches the smaller circle.

The perpendicular from the centre to the chord is the radius of the smaller circle, $3$ cm, and the radius of the larger circle, $5$ cm, is the hypotenuse. So half the chord is

$$\sqrt{5^2-3^2}=\sqrt{25-9}=\sqrt{16}=4\text{ cm}$$

and the whole chord is

$$2\times 4=8\text{ cm}$$

The doubling at the end is the step that gets forgotten. The right triangle only contains half the chord, because the perpendicular from the centre lands at the mid-point. **Answering $4$ cm is the single most common error in this type of question, and the guard is to ask whether the answer is a half-chord or a whole chord before writing it down.

Result 4 — opposite sides of a circumscribed quadrilateral subtend supplementary angles at the centre.** In quadrilateral $ABCD$ circumscribing a circle with centre $O$,

$$\angle AOB+\angle COD=180^{\circ }$$

Why. Joining the centre to the four vertices and to the four points of contact splits the figure into eight triangles, congruent in pairs by the equal-tangent result. The eight angles at $O$ therefore come in four equal pairs and sum to $360^{\circ }$. Grouping them shows that the two angles named use exactly half of that total, giving $180^{\circ }$.

Worked example 4 — a tangent through a point of intersection. Two tangents are drawn from an external point $P$ to a circle, touching at $A$ and $B$. Prove that $OP$ is the perpendicular bisector of $AB$.

Since $PA=PB$ and $OA=OB$, both $P$ and $O$ are equidistant from $A$ and $B$. Two distinct points equidistant from the ends of a segment determine its perpendicular bisector, so the line $OP$ is the perpendicular bisector of $AB$. Two lines of reasoning, no construction needed — and it is the tidiest proof in the chapter.