Circles

The chapter is compact — just two numbered theorems — but each is load-bearing. Section 10.2, "Tangent to a Circle," sorts every possible line-circle relationship into three cases (non-intersecting, secant, tangent), defines the point of contact, and proves Theorem 10.1: the tangent at any point is perpendicular to the radius through that point. This is why "join the centre to the point of contact" is the automatic first move in almost any tangent proof. Section 10.3, "Number of Tangents from a Point on a Circle," classifies points by location (inside, on, or outside the circle) and proves Theorem 10.2: the two tangents from an external point have equal length. NCERT's own worked examples build straight on these — Example 1 uses Theorem 10.1 to show a chord of a larger circle tangent to a smaller concentric circle is bisected at the point of contact; Example 2 uses Theorem 10.2 and angle-sum reasoning to relate the angle between two tangents to the angle at the centre; Example 3 combines both with similar triangles to find an unknown tangent length. The exercises extend this two-theorem toolkit to quadrilaterals and triangles circumscribing a circle, where most of the chapter's board-exam value lives.

Tangent, secant, and non-intersecting line — the three possibilities

Given a circle and a line in the same plane, exactly three things can happen, and no fourth possibility exists. If they share no common point, the line is a non-intersecting line with respect to the circle. If they share exactly two common points, the line is a secant of the circle. If they share exactly one common point, the line is a tangent, and that single shared point is the point of contact. A tangent is best understood as a secant pushed to its limit: slide a secant's two intersection points toward each other along the circle, and the moment they merge into one, the secant has become a tangent. This is why NCERT calls the tangent "a special case of the secant, when the two end points of its corresponding chord coincide."

Theorem 10.1 — the tangent is perpendicular to the radius at the point of contact

The rolling-wheel picture from the overview is exactly the content of the chapter's first theorem, stated formally:

where: $O$ is the centre of the circle, $P$ is the point of contact, and $XY$ is the tangent line at $P$. The full step-by-step proof appears in the derivation block below; the underlying idea is that $P$ is the closest point of the line $XY$ to the centre $O$ — every other point of the tangent line lies strictly outside the circle and is therefore strictly farther from $O$ than the radius $OP$. A line segment from a point to a line is shortest exactly when it is perpendicular to that line, which forces $OP \perp XY$. Two useful remarks follow immediately: there is exactly one tangent at each point of a circle (since perpendicularity to a given radius fixes the tangent line's direction uniquely), and the radius drawn to the point of contact is sometimes called the normal to the circle at that point.

Number of tangents from a point, and the length of a tangent

Where a point sits relative to a circle decides how many tangents can be drawn from it. If $P$ lies inside the circle, every line through $P$ meets the circle at two points, so no tangent exists. If $P$ lies on the circle, Theorem 10.1's uniqueness remark gives exactly one tangent. If $P$ lies outside, exactly two tangents can be drawn, touching the circle at $T_1$ and $T_2$. The segment $PT_1$ (or $PT_2$) is called the length of the tangent from $P$. NCERT's Theorem 10.2 says these two lengths are always equal:

where: $P$ is any point outside the circle, and $T_1$, $T_2$ are the points of contact of the two tangents drawn from $P$. Combining this with Theorem 10.1's right angle at each point of contact gives a formula used constantly in numerical problems: in right triangle $OT_1P$, with $OT_1 = r$ (the radius) and $OP = d$ (the distance from the external point to the centre), Pythagoras gives the tangent length directly,

where: $d$ is the straight-line distance from the external point $P$ to the centre $O$, and $r$ is the radius; the condition $d > r$ is exactly the statement that $P$ lies outside the circle, which is what guarantees a tangent (and a positive length under the square root) exists at all.

Angle relations and circumscribing polygons

Two further results appear in almost every exercise problem. First, since $OP$ bisects the angle between the two tangents (a remark attached to Theorem 10.2's proof), the angle between the two tangents and the angle the two radii make at the centre are always supplementary:

where: $\angle T_1PT_2$ is the angle between the two tangents at the external point, and $\angle T_1OT_2$ is the angle subtended at the centre by the two radii to the points of contact; the full proof is given as the third derivation below. Second, when a quadrilateral $ABCD$ is drawn so that each of its four sides touches a common circle (the quadrilateral circumscribes the circle), the equal-tangent-length property applied at each of the four vertices in turn forces

where: each side is broken by its point of contact into two tangent segments from the two vertices at its ends, and equal tangent lengths from each shared vertex let all eight segments be paired and regrouped into this single sum identity. This identity alone proves a parallelogram circumscribing a circle must be a rhombus: opposite sides of a parallelogram are already equal, so $AB+CD=AD+BC$ with $AB=CD$ and $AD=BC$ collapses to $AB=AD$ — all four sides equal.

Board level - the chord of a larger circle tangent to a smaller concentric circle (extends NCERT Example 1)

Two concentric circles have radii $5\ \text{cm}$ and $3\ \text{cm}$. Find the length of the chord of the larger circle that touches the smaller circle.

Step 1. Let $O$ be the common centre, and let $AB$ be the chord of the larger circle (radius $OA = 5\ \text{cm}$) that touches the smaller circle (radius $3\ \text{cm}$) at the point $P$. Since $AB$ is tangent to the smaller circle at $P$, Theorem 10.1 gives $OP \perp AB$, with $OP = 3\ \text{cm}$.

Step 2. Since $OP \perp AB$ and $OP$ is a radius of the larger circle drawn to the chord $AB$, the perpendicular from the centre bisects the chord (a Class IX result), so $AP = PB$, and $AB = 2\,AP$.

Step 3. In right triangle $OPA$, apply Pythagoras with hypotenuse $OA = 5\ \text{cm}$ (a radius of the larger circle) and leg $OP = 3\ \text{cm}$: $AP = \sqrt{OA^2 - OP^2} = \sqrt{5^2 - 3^2} = \sqrt{25-9} = \sqrt{16} = 4\ \text{cm}$.

Step 4. Double $AP$ to get the full chord: $AB = 2 \times 4 = 8\ \text{cm}$. Answer: the chord has length $\mathbf{8\ \text{cm}}$.

Board level - tangent length, radius, and the Pythagorean formula (extends Exercise 10.2, Q.1)

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

Step 1. Let $O$ be the centre and $T$ the point of contact of the tangent from $Q$. By Theorem 10.1, $OT \perp QT$, so $\triangle OTQ$ is right-angled at $T$, with hypotenuse $OQ = 25\ \text{cm}$ and one leg $QT = 24\ \text{cm}$ (the given tangent length).

Step 2. Apply Pythagoras to find the missing leg $OT$, which is the radius: $OT^2 = OQ^2 - QT^2 = 25^2 - 24^2 = 625 - 576 = 49$.

Step 3. Take the square root: $OT = \sqrt{49} = 7\ \text{cm}$. Answer: the radius of the circle is $\mathbf{7\ \text{cm}}$.

Board HOTS / foundation JEE-NEET - a triangle circumscribing a circle (extends Exercise 10.2, Q.12)

A triangle $ABC$ is drawn to circumscribe a circle of radius $4\ \text{cm}$, such that the point of contact $D$ on side $BC$ divides it into segments $BD = 8\ \text{cm}$ and $DC = 6\ \text{cm}$. Find the lengths of sides $AB$ and $AC$.

Step 1. Let the circle touch $BC$, $CA$, $AB$ at $D$, $E$, $F$. By Theorem 10.2, tangents from the same vertex are equal, so $BD=BF=8\ \text{cm}$, $CD=CE=6\ \text{cm}$, and $AE=AF=x\ \text{cm}$ (unknown). This gives $AB = x+8$ and $AC = x+6$.

Step 2. Set up the area two ways. With sides $a=BC=14$, $b=CA=x+6$, $c=AB=x+8$, the semi-perimeter is $s = \dfrac{a+b+c}{2} = x+14$.

Step 3. Split $\triangle ABC$ into three triangles meeting at the incentre $O$, each with height equal to the inradius $r=4\ \text{cm}$ (since $OD,OE,OF \perp$ the sides by Theorem 10.1). So $\text{ar}(\triangle ABC) = \tfrac12 r(a+b+c) = \tfrac12 \times 4 \times 2s = 4s = 4(x+14)$.

Step 4. Equate this to Heron's formula, $\text{ar}(\triangle ABC) = \sqrt{s(s-a)(s-b)(s-c)}$. With $s = x+14$: $s-a = (x+14)-14 = x$; $s-b = (x+14)-(x+6) = 8$; $s-c = (x+14)-(x+8) = 6$. So $4(x+14) = \sqrt{(x+14)\cdot x \cdot 8 \cdot 6} = \sqrt{48x(x+14)}$.

Step 5. Square both sides: $16(x+14)^2 = 48x(x+14)$. Divide through by the common factor $(x+14)$, valid since $x+14 \ne 0$: $16(x+14) = 48x$, i.e. $16x + 224 = 48x$, so $224 = 32x$, giving $x = 7\ \text{cm}$. So $AE = AF = 7\ \text{cm}$.

Step 6. Substitute back into Step 1: $AB = x+8 = 7+8 = 15\ \text{cm}$ and $AC = x+6 = 7+6 = 13\ \text{cm}$. Answer: $AB = \mathbf{15\ \text{cm}}$ and $AC = \mathbf{13\ \text{cm}}$. Timing note: budget 6 to 7 minutes — this is the chapter's most demanding HOTS pattern, chaining Theorem 10.2's tangent-equality onto the area-via-inradius identity $\text{ar}=rs$, a technique that reaches slightly beyond this chapter's own two theorems but is what examiners expect for circumscribing-triangle problems. The reusable skeleton: name the three unknown tangent segments using Theorem 10.2, express the three sides in terms of them, then bring in $\text{ar}=rs$ (or Heron's formula, or a right-angle condition) to pin down the last unknown.

The most common error is skipping the first move: not joining the radius to the point of contact, so Theorem 10.1's right angle — the fact that unlocks Pythagoras — never enters the solution. A second mistake is confusing a secant with a tangent in a figure that looks like it grazes the circle; check explicitly whether one or two intersection points are implied. A third trap is writing $PT=\sqrt{r^2-d^2}$ instead of $PT=\sqrt{d^2-r^2}$ — $d$ (distance from the external point to the centre) is always the larger quantity, since $P$ lies outside the circle. A fourth is assuming any two tangents to a circle are equal regardless of where they are drawn from — Theorem 10.2 only guarantees equality for tangents from the same external point. A fifth, common in circumscribing-polygon problems, is forgetting to name the tangent lengths from each vertex as equal before setting up the perimeter or area equation. Finally, students mix up "angle between the tangents" ($\angle QPR$) with "angle at the centre" ($\angle QOR$) and treat them as equal instead of supplementary — the correct relation is $\angle QPR + \angle QOR = 180°$.

• State precisely the difference between a tangent, a secant, and a non-intersecting line, and identify which is which from a figure.
• State Theorem 10.1 (tangent perpendicular to the radius at the point of contact) and reproduce its shortest-distance proof.
• Explain why there is exactly one tangent at each point of a circle, and why the radius to that point is sometimes called the normal.
• Classify the number of tangents from a point as zero, one, or two, based on whether the point lies inside, on, or outside the circle.
• State Theorem 10.2 (equal tangent lengths from an external point) and reproduce its RHS-congruence proof, plus the Pythagorean shortcut.
• Compute the length of a tangent from an external point given the radius and the distance to the centre, using $PT=\sqrt{d^2-r^2}$.
• Prove that the angle between two tangents from an external point is supplementary to the angle the two radii make at the centre.
• Solve quadrilateral- and triangle-circumscribing-a-circle problems by naming equal tangent segments from each vertex first.
• Recognise and avoid the chapter's classic traps: skipping the radius, confusing $d$ and $r$ in the Pythagorean formula, and mixing up angle equality with angle supplementarity.

Ask the tutor: Want to work through a circumscribing-quadrilateral or circumscribing-triangle proof from your own textbook, step by step, using this chapter's two theorems?