Tangent to a circle

Look at a bicycle wheel resting against a straight kerb, or a ball rolling to a stop against a wall — at the point of contact, the wheel's rim and the straight edge touch at exactly one point and nowhere else. That single-point touch is the geometric idea behind a tangent to a circle, the opening concept of the Circles chapter in CBSE Class 10 Mathematics. A tangent is not just "a line near a circle" — it is the precise limiting case of a secant, the line that normally cuts a circle at two points, as those two points slide together into one. That limiting picture hands us the single most useful fact in this chapter: the tangent at any point is always perpendicular to the radius drawn to that point. This one fact turns almost every tangent problem into a right-triangle problem, solvable with the Pythagoras theorem you already know well.

The core insight

A tangent is the limiting position of a secant whose two points of intersection merge into one, and at that point of contact the tangent is always perpendicular to the radius, which turns tangent-length problems into applications of the Pythagoras theorem.

The core idea

A secant is a line that cuts a circle at two distinct points. A chord is the segment of that secant lying inside the circle, joining those two points. A tangent, by contrast, touches the circle at exactly one point — the point of contact — and does not enter the interior anywhere else.

Picture sliding one of a secant's two intersection points along the circle toward the other. As the gap closes to zero, the line settles into the tangent at that point — the tangent is the limiting position of a secant.
Through any point on the circle, exactly one tangent can be drawn.
Through any point outside the circle, exactly two tangents can be drawn, and (a fact used constantly in problems) both tangent segments from that external point are equal in length.

The single fact that makes tangent geometry tractable links the tangent to the circle's centre:

OP \perp XY, \qquad \text{i.e., } \angle OPX = \angle OPY = 90^\circ.

where: $O$ is the centre of the circle, $P$ is the point of contact, and $XY$ is the tangent line drawn at $P$ . In words: the tangent at any point of a circle is perpendicular to the radius through the point of contact. (This is proved by contradiction in the companion article on the tangent theorem; here it is the working tool.) Because the radius meets the tangent at a right angle, joining the centre $O$ to any external point $A$ from which a tangent touches the circle at $P$ always produces a right triangle $OPA$ , right-angled at $P$ , with $OA$ as the hypotenuse — the shape every tangent-length calculation reduces to.

Figure

A circle with centre O, radius OP drawn to the point of contact P, the tangent line through P perpendicular to OP, and an external point A joined to both O and P forming a right triangle with the right angle at P

Derivation ATangent-length formula

The goal of this derivation is to turn the perpendicularity fact into a usable formula: given the radius of a circle and the distance from an external point to the centre, find the length of the tangent drawn from that external point, using nothing beyond the Pythagoras theorem applied to the right triangle formed by the centre, the point of contact, and the external point.

Step 1. Let $O$ be the centre of a circle of radius $r$ , and let $A$ be a point outside the circle at distance $OA = d$ from the centre. Let a tangent from $A$ touch the circle at $P$ , so $AP$ is the tangent length we want.

Step 2. By the tangent–radius theorem, $OP \perp AP$ , so $\angle OPA = 90^\circ$ . Triangle $OPA$ is therefore right-angled at $P$ , and the side $OA$ , joining the centre to the external point, is the hypotenuse.

Step 3. Apply the Pythagoras theorem to triangle $OPA$ :

OA^2 = OP^2 + AP^2 \implies d^2 = r^2 + AP^2.

Step 4. Solve for the tangent length:

AP = \sqrt{d^2 - r^2}.

where: $AP$ is the length of the tangent from the external point, $d = OA$ is the distance from the external point to the centre, and $r$ is the radius. Note the formula only makes sense when $d > r$ (the point really is outside the circle); if $d = r$ the point lies on the circle and no tangent segment exists, and if $d < r$ the point is inside, where no tangent can be drawn at all.

Worked example

Problem. From an external point $A$ , a tangent $AP$ is drawn to a circle of radius $7\text{ cm}$ with centre $O$ , touching the circle at $P$ . If $OA = 25\text{ cm}$ , find the length of the tangent $AP$ . A second tangent $AQ$ is drawn from the same point $A$ , touching the circle at $Q$ — find $AQ$ as well.

Step 1. Since $P$ is the point of contact, the radius $OP$ is perpendicular to the tangent $AP$ : $\angle OPA = 90^\circ$ , so triangle $OPA$ is right-angled at $P$ with hypotenuse $OA$ .

Step 2. Apply the tangent-length formula with $d = OA = 25\text{ cm}$ and $r = OP = 7\text{ cm}$ :

AP = \sqrt{d^2 - r^2} = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576}.

Step 3. Take the square root: $AP = 24\text{ cm}$ .

Step 4. For the second tangent $AQ$ from the same external point $A$ : since tangent segments drawn from a single external point to a circle are always equal in length, $AQ = AP$ without any further calculation.

Answer: $AP = 24\text{ cm}$ and $AQ = 24\text{ cm}$ .

Competitive stretch

If instead the tangent length is given as $AP = 24\text{ cm}$ and the radius is $r = 7\text{ cm}$ , the same right triangle gives $OA^2 = AP^2 + OP^2 = 24^2 + 7^2 = 576 + 49 = 625$ , so $OA = 25\text{ cm}$ — the formula runs equally well in reverse to find the distance to the centre when the tangent length and radius are known.

Common traps

Confusing a tangent with a chord or secant. A chord or secant meets the circle at two points; a tangent meets it at exactly one.
Forgetting which side is the hypotenuse. In triangle $OPA$ , the segment from the centre to the external point ( $OA$ ) is always the hypotenuse — never the radius or the tangent segment.
Applying the formula when the point is not outside the circle. $AP = \sqrt{d^2 - r^2}$ needs $d > r$ ; check this before computing, especially in word problems that don't state it directly.
Assuming any two tangent segments from different external points are equal. Equal tangent lengths only hold for two tangents drawn from the same external point — not between unrelated points.

Ask the tutor: Want to see the full proof of why the tangent is exactly perpendicular to the radius, or try a few more mixed tangent-length problems? Ask the AI tutor to walk through them with you.

How is a tangent different from a chord that just looks very short?

A chord always joins two distinct points on the circle, however close together they look on a rough diagram — it still passes through the interior between them. A tangent, by contrast, has genuinely only one point in common with the circle; it never dips inside. However close a chord's endpoints appear, drawing it accurately will show two intersection points, while a true tangent will show only one, however far you extend it.

Why is OA always the hypotenuse and never the radius or the tangent?

The right angle in the triangle sits at $P$ , the point of contact, because that is where the tangent meets the radius perpendicularly. The hypotenuse is always the side opposite the right angle, which is the side that does not touch $P$ — that is $OA$ , the segment from the centre to the external point. Since $OP$ and $AP$ both meet at $P$ , neither of them can be the hypotenuse.

Do I need to prove the tangent-radius perpendicularity every time I use it?

No. In CBSE Class 10, once the theorem "the tangent at any point of a circle is perpendicular to the radius through the point of contact" has been established, you are expected to use it directly in problems, citing it by name, rather than re-proving it each time. Save the full proof-by-contradiction argument for questions that explicitly ask you to prove the theorem.

Tangent to a circle

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads