Class 10 › Mathematics › Chapter
Circles
Tangents, secants, and fundamental geometric proofs governing the properties of circles.
1. Circle Definition
A circle is the locus of all points in a plane that are at a fixed distance from a fixed point in that plane.
Centre: The fixed point is called the centre ($O$).
Radius: The fixed constant distance from the centre to any point on the boundary is called the radius ($r$).
Circle $C(O, r)$ with centre $O$ and radius $OP = r$.
2. Secant
A secant is a straight line that intersects a circle at two distinct points. While a chord is a finite segment whose endpoints lie on the circle, a secant is an extended straight line containing that chord.
Line $l$ cuts the circle at two distinct points $A$ and $B$.
3. Tangent to a Circle
A tangent to a circle is a straight line that intersects the circle at only one point. The common point of the circle and the tangent is called the point of contact.
Limiting Case: A tangent can be viewed as the limiting case of a secant when its two points of intersection coincide.
4. Number of Tangents to a Circle
The position of a point relative to a circle strictly determines the number of tangents that can be drawn from it:
- Case 1 (Inside the circle): No tangent can be drawn to a circle through a point lying inside it ($0$ tangents). Any line through an interior point is a secant.
- Case 2 (On the circle): There is one and only one tangent to a circle passing through a point lying on the circle.
- Case 3 (Outside the circle): There are exactly two tangents that can be drawn to a circle from an external point.
From an external point $P$, exactly two tangents $PT_1$ and $PT_2$ exist.
5. Length of Tangent
The length of the segment of the tangent from an external point to the point of contact with the circle is called the length of the tangent.
Relation
If $P$ is an external point at distance $d$ from centre $O$, and $r$ is the radius, then length $L = \sqrt{d^2 - r^2}$.
6. Theorem 1
Statement: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Tangent $XY$ touching at $P$; $Q$ is any point on $XY$ other than $P$.
Proof
Given: A circle with centre $O$ and a tangent $XY$ touching the circle at point $P$.
To Prove: $OP \perp XY$.
Construction: Take any point $Q$ on $XY$ other than $P$. Join $OQ$. Let $OQ$ intersect the circle at $R$.
Proof:
1. Point $Q$ lies outside the circle (if $Q$ lay inside, $XY$ would become a secant and not a tangent).
2. Therefore, $OQ > \text{radius } OP$.
3. In fact, $OQ = OR + RQ$. Since $OR = OP$ (radii of the same circle), we have: $$OQ = OP + RQ \implies OQ > OP$$ 4. This holds true for every point on the line $XY$ except point $P$.
5. Hence, $OP$ is the shortest distance from the centre $O$ to the line $XY$.
6. Since the shortest distance from a point to a straight line is the perpendicular distance, $OP \perp XY$.
Hence proved.
1. Point $Q$ lies outside the circle (if $Q$ lay inside, $XY$ would become a secant and not a tangent).
2. Therefore, $OQ > \text{radius } OP$.
3. In fact, $OQ = OR + RQ$. Since $OR = OP$ (radii of the same circle), we have: $$OQ = OP + RQ \implies OQ > OP$$ 4. This holds true for every point on the line $XY$ except point $P$.
5. Hence, $OP$ is the shortest distance from the centre $O$ to the line $XY$.
6. Since the shortest distance from a point to a straight line is the perpendicular distance, $OP \perp XY$.
Hence proved.
7. Theorem 2 (Converse of Theorem 1)
Statement: A line drawn through the end of a radius and perpendicular to it is a tangent to the circle.
Line $AB \perp OP$ at outer end $P$.
Proof
Given: A circle with centre $O$, radius $OP$, and a line $AB$ passing through $P$ such that $AB \perp OP$.
To Prove: $AB$ is a tangent to the circle at point $P$.
Construction: Take any point $Q$ on $AB$ distinct from $P$. Join $OQ$.
Proof:
1. In $\triangle OPQ$, $\angle OPQ = 90^\circ$ (given $AB \perp OP$).
2. In a right-angled triangle, the hypotenuse is the strictly longest side. Thus: $$OQ > OP$$ 3. Since $OP$ is the radius of the circle, $OQ > \text{radius}$.
4. Therefore, point $Q$ lies outside the circle.
5. Since this is true for every point on $AB$ except $P$, $P$ is the only point of $AB$ that lies on the circle.
6. A line having only one point in common with a circle is a tangent. Hence, $AB$ is tangent to the circle at $P$.
Hence proved.
1. In $\triangle OPQ$, $\angle OPQ = 90^\circ$ (given $AB \perp OP$).
2. In a right-angled triangle, the hypotenuse is the strictly longest side. Thus: $$OQ > OP$$ 3. Since $OP$ is the radius of the circle, $OQ > \text{radius}$.
4. Therefore, point $Q$ lies outside the circle.
5. Since this is true for every point on $AB$ except $P$, $P$ is the only point of $AB$ that lies on the circle.
6. A line having only one point in common with a circle is a tangent. Hence, $AB$ is tangent to the circle at $P$.
Hence proved.
8. Theorem 3
Statement: The lengths of tangents drawn from an external point to a circle are equal.
Tangents $PQ$ and $PR$ drawn from external point $P$.
Proof
Given: A circle with centre $O$, an external point $P$, and two tangents $PQ$ and $PR$ touching the circle at $Q$ and $R$ respectively.
To Prove: $PQ = PR$.
Construction: Join $OP$, $OQ$, and $OR$.
Proof:
1. Since tangent is perpendicular to radius at the point of contact (Theorem 1): $$\angle OQP = 90^\circ \quad \text{and} \quad \angle ORP = 90^\circ$$ 2. In right triangles $\triangle OQP$ and $\triangle ORP$: $$\begin{aligned} \angle OQP &= \angle ORP = 90^\circ && \text{(Each } 90^\circ\text{)} \\ OP &= OP && \text{(Common hypotenuse)} \\ OQ &= OR && \text{(Radii of the same circle)} \end{aligned}$$ 3. By the **RHS congruence criterion**: $$\triangle OQP \cong \triangle ORP$$ 4. By corresponding parts of congruent triangles (CPCT): $$PQ = PR$$ Hence proved.
1. Since tangent is perpendicular to radius at the point of contact (Theorem 1): $$\angle OQP = 90^\circ \quad \text{and} \quad \angle ORP = 90^\circ$$ 2. In right triangles $\triangle OQP$ and $\triangle ORP$: $$\begin{aligned} \angle OQP &= \angle ORP = 90^\circ && \text{(Each } 90^\circ\text{)} \\ OP &= OP && \text{(Common hypotenuse)} \\ OQ &= OR && \text{(Radii of the same circle)} \end{aligned}$$ 3. By the **RHS congruence criterion**: $$\triangle OQP \cong \triangle ORP$$ 4. By corresponding parts of congruent triangles (CPCT): $$PQ = PR$$ Hence proved.
9. Theorem 4
Statement: If two tangents are drawn from an external point to a circle:
(i) they subtend equal angles at the centre, and
(ii) they are equally inclined to the line segment joining the centre to that point.
Angles subtended at the centre ($\angle 1, \angle 2$) and at $P$ ($\angle 3, \angle 4$).
Proof
Given: $PQ$ and $PR$ are two tangents drawn from external point $P$ to a circle with centre $O$.
To Prove:
(i) $\angle QOP = \angle ROP$ (they subtend equal angles at the centre)
(ii) $\angle QPO = \angle RPO$ (equally inclined to line $OP$)
(i) $\angle QOP = \angle ROP$ (they subtend equal angles at the centre)
(ii) $\angle QPO = \angle RPO$ (equally inclined to line $OP$)
Proof:
1. From Theorem 3, consider $\triangle OQP$ and $\triangle ORP$.
2. We have: $$\begin{aligned} OQ &= OR && \text{(Radii)} \\ PQ &= PR && \text{(Tangents from external point)} \\ OP &= OP && \text{(Common side)} \end{aligned}$$ 3. By the **SSS congruence criterion** (or RHS): $$\triangle OQP \cong \triangle ORP$$ 4. Corresponding angles are equal (CPCT): $$\angle QOP = \angle ROP \quad \text{and} \quad \angle QPO = \angle RPO$$ Hence proved.
1. From Theorem 3, consider $\triangle OQP$ and $\triangle ORP$.
2. We have: $$\begin{aligned} OQ &= OR && \text{(Radii)} \\ PQ &= PR && \text{(Tangents from external point)} \\ OP &= OP && \text{(Common side)} \end{aligned}$$ 3. By the **SSS congruence criterion** (or RHS): $$\triangle OQP \cong \triangle ORP$$ 4. Corresponding angles are equal (CPCT): $$\angle QOP = \angle ROP \quad \text{and} \quad \angle QPO = \angle RPO$$ Hence proved.
10. Theorem 5
Statement: Prove that in two concentric circles, the chord of the larger circle which touches the smaller circle is bisected at the point of contact.
Chord $AB$ of outer circle $C_1$ touches inner circle $C_2$ at $P$.
Proof
Given: Two concentric circles $C_1$ and $C_2$ with common centre $O$. $AB$ is a chord of the larger circle $C_1$ touching the smaller circle $C_2$ at point $P$.
To Prove: $AP = PB$ (chord $AB$ is bisected at $P$).
Construction: Join $OP$.
Proof:
1. Since $AB$ is a tangent to the smaller circle $C_2$ at point $P$, and $OP$ is the radius through the point of contact: $$OP \perp AB \quad \text{(Theorem 1)}$$ 2. Now consider $AB$ as a chord of the larger circle $C_1$.
3. We know that the perpendicular drawn from the centre of a circle to a chord bisects the chord.
4. Since $OP \perp AB$ and $OP$ passes through centre $O$: $$AP = PB$$ Hence proved.
1. Since $AB$ is a tangent to the smaller circle $C_2$ at point $P$, and $OP$ is the radius through the point of contact: $$OP \perp AB \quad \text{(Theorem 1)}$$ 2. Now consider $AB$ as a chord of the larger circle $C_1$.
3. We know that the perpendicular drawn from the centre of a circle to a chord bisects the chord.
4. Since $OP \perp AB$ and $OP$ passes through centre $O$: $$AP = PB$$ Hence proved.
11. Theorem 6
Statement: Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Tangents $PQ$ and $RS$ drawn at diameter endpoints $A$ and $B$.
Proof
Given: A circle with centre $O$ and diameter $AB$. Lines $PQ$ and $RS$ are tangents to the circle at endpoints $A$ and $B$ respectively.
To Prove: $PQ \parallel RS$.
Proof:
1. Since $OA$ is the radius to the point of contact $A$ of tangent $PQ$: $$OA \perp PQ \implies \angle PAB = 90^\circ \quad \text{--- (1)}$$ 2. Similarly, $OB$ is the radius to the point of contact $B$ of tangent $RS$: $$OB \perp RS \implies \angle ABS = 90^\circ \quad \text{--- (2)}$$ 3. From (1) and (2): $$\angle PAB = \angle ABS = 90^\circ$$ 4. But these are a pair of alternate interior angles formed by the transversal line $AB$ intersecting lines $PQ$ and $RS$.
5. When alternate interior angles are equal, the lines are parallel. $$\therefore PQ \parallel RS$$ Hence proved.
1. Since $OA$ is the radius to the point of contact $A$ of tangent $PQ$: $$OA \perp PQ \implies \angle PAB = 90^\circ \quad \text{--- (1)}$$ 2. Similarly, $OB$ is the radius to the point of contact $B$ of tangent $RS$: $$OB \perp RS \implies \angle ABS = 90^\circ \quad \text{--- (2)}$$ 3. From (1) and (2): $$\angle PAB = \angle ABS = 90^\circ$$ 4. But these are a pair of alternate interior angles formed by the transversal line $AB$ intersecting lines $PQ$ and $RS$.
5. When alternate interior angles are equal, the lines are parallel. $$\therefore PQ \parallel RS$$ Hence proved.
12. Theorem 7
Statement: Prove that the line segment joining the points of contact of two parallel tangents to a circle is a diameter of the circle.
Parallel tangents $l \parallel m$ touching at $A$ and $B$.
Proof
Given: A circle with centre $O$. Tangents $l$ and $m$ are parallel, touching the circle at $A$ and $B$ respectively.
To Prove: $AOB$ is a straight line passing through centre $O$ (i.e., $AB$ is a diameter).
Construction: Draw a line $XOY \parallel l$ passing through centre $O$.
Proof:
1. Since $l$ is a tangent at $A$ and $OA$ is the radius: $$OA \perp l \implies \angle OAP = 90^\circ$$ 2. Since $l \parallel XOY$, the consecutive interior angles on the same side of transversal $OA$ add up to $180^\circ$: $$\angle OAP + \angle AOY = 180^\circ \implies 90^\circ + \angle AOY = 180^\circ \implies \angle AOY = 90^\circ \quad \text{--- (1)}$$ 3. Since $l \parallel m$ and $XOY \parallel l$, we have $XOY \parallel m$.
4. Since $m$ is tangent at $B$ and $OB$ is the radius, $OB \perp m \implies \angle OBR = 90^\circ$.
5. Again, by consecutive interior angles for $XOY \parallel m$ cut by transversal $OB$: $$\angle OBR + \angle BOY = 180^\circ \implies 90^\circ + \angle BOY = 180^\circ \implies \angle BOY = 90^\circ \quad \text{--- (2)}$$ 6. Adding equations (1) and (2): $$\angle AOY + \angle BOY = 90^\circ + 90^\circ = 180^\circ$$ 7. Therefore, $\angle AOB$ is a straight angle ($180^\circ$), which proves that points $A, O, B$ are collinear.
8. Since $A$ and $B$ lie on the circle and the line segment passes through centre $O$, $AB$ is a diameter.
Hence proved.
1. Since $l$ is a tangent at $A$ and $OA$ is the radius: $$OA \perp l \implies \angle OAP = 90^\circ$$ 2. Since $l \parallel XOY$, the consecutive interior angles on the same side of transversal $OA$ add up to $180^\circ$: $$\angle OAP + \angle AOY = 180^\circ \implies 90^\circ + \angle AOY = 180^\circ \implies \angle AOY = 90^\circ \quad \text{--- (1)}$$ 3. Since $l \parallel m$ and $XOY \parallel l$, we have $XOY \parallel m$.
4. Since $m$ is tangent at $B$ and $OB$ is the radius, $OB \perp m \implies \angle OBR = 90^\circ$.
5. Again, by consecutive interior angles for $XOY \parallel m$ cut by transversal $OB$: $$\angle OBR + \angle BOY = 180^\circ \implies 90^\circ + \angle BOY = 180^\circ \implies \angle BOY = 90^\circ \quad \text{--- (2)}$$ 6. Adding equations (1) and (2): $$\angle AOY + \angle BOY = 90^\circ + 90^\circ = 180^\circ$$ 7. Therefore, $\angle AOB$ is a straight angle ($180^\circ$), which proves that points $A, O, B$ are collinear.
8. Since $A$ and $B$ lie on the circle and the line segment passes through centre $O$, $AB$ is a diameter.
Hence proved.
13. Theorem 8
Statement: Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Angle $\angle APB$ between tangents is supplementary to central angle $\angle AOB$.
Proof
Given: $PA$ and $PB$ are two tangents drawn from external point $P$ to a circle with centre $O$, touching at $A$ and $B$.
To Prove: $\angle APB + \angle AOB = 180^\circ$.
Proof:
1. Since the tangent at any point is perpendicular to the radius through the point of contact (Theorem 1): $$OA \perp PA \implies \angle OAP = 90^\circ$$ $$OB \perp PB \implies \angle OBP = 90^\circ$$ 2. Consider the quadrilateral $OAPB$. The sum of all four interior angles of a quadrilateral is $360^\circ$: $$\angle APB + \angle OAP + \angle AOB + \angle OBP = 360^\circ$$ 3. Substitute the values of $\angle OAP$ and $\angle OBP$: $$\angle APB + 90^\circ + \angle AOB + 90^\circ = 360^\circ$$ $$\angle APB + \angle AOB + 180^\circ = 360^\circ$$ $$\angle APB + \angle AOB = 360^\circ - 180^\circ = 180^\circ$$ 4. Thus, $\angle APB$ and $\angle AOB$ are supplementary angles.
Hence proved.
1. Since the tangent at any point is perpendicular to the radius through the point of contact (Theorem 1): $$OA \perp PA \implies \angle OAP = 90^\circ$$ $$OB \perp PB \implies \angle OBP = 90^\circ$$ 2. Consider the quadrilateral $OAPB$. The sum of all four interior angles of a quadrilateral is $360^\circ$: $$\angle APB + \angle OAP + \angle AOB + \angle OBP = 360^\circ$$ 3. Substitute the values of $\angle OAP$ and $\angle OBP$: $$\angle APB + 90^\circ + \angle AOB + 90^\circ = 360^\circ$$ $$\angle APB + \angle AOB + 180^\circ = 360^\circ$$ $$\angle APB + \angle AOB = 360^\circ - 180^\circ = 180^\circ$$ 4. Thus, $\angle APB$ and $\angle AOB$ are supplementary angles.
Hence proved.
14. Theorem 9
Statement: Prove that there is one and only one tangent at any point on the circumference of a circle.
Uniqueness of the perpendicular line through point $P$ on radius $OP$.
Proof
Given: A circle with centre $O$ and a point $P$ on its circumference.
To Prove: Exactly one tangent can be drawn to the circle at point $P$.
Proof:
1. Join $OP$. $OP$ is the unique radius to the point of contact $P$.
2. By Theorem 1, any tangent to the circle at $P$ must be perpendicular to $OP$.
3. From planar Euclidean geometry, at a given point on a line, there exists one and only one line perpendicular to the given line in that plane.
4. Therefore, only one line $t$ can be drawn through point $P$ such that $t \perp OP$.
5. By Theorem 2 (Converse), this unique perpendicular line is a tangent to the circle.
6. Hence, there is one and only one tangent at any point on the circumference of a circle.
Hence proved.
1. Join $OP$. $OP$ is the unique radius to the point of contact $P$.
2. By Theorem 1, any tangent to the circle at $P$ must be perpendicular to $OP$.
3. From planar Euclidean geometry, at a given point on a line, there exists one and only one line perpendicular to the given line in that plane.
4. Therefore, only one line $t$ can be drawn through point $P$ such that $t \perp OP$.
5. By Theorem 2 (Converse), this unique perpendicular line is a tangent to the circle.
6. Hence, there is one and only one tangent at any point on the circumference of a circle.
Hence proved.
15. Theorem 10
Statement: Prove that the perpendicular at the point of contact of the tangent to a circle passes through the centre.
Proof by contradiction: assuming perpendicular $O'P$ does not pass through centre $O$.
Proof (By Contradiction)
Given: A tangent $AB$ to a circle with centre $O$, touching at point $P$. Line $PQ \perp AB$ at point $P$.
To Prove: The perpendicular $PQ$ passes through the centre $O$.
Proof:
1. Suppose, if possible, that the perpendicular to $AB$ at $P$ does not pass through the centre $O$.
2. Then let it pass through some other point $O'$. Join $O'P$.
3. By our assumption, $O'P \perp AB$: $$\angle O'PB = 90^\circ \quad \text{--- (1)}$$ 4. But we know from Theorem 1 that the line joining the centre $O$ to the point of contact $P$ is perpendicular to the tangent $AB$: $$OP \perp AB \implies \angle OPB = 90^\circ \quad \text{--- (2)}$$ 5. Comparing equations (1) and (2): $$\angle O'PB = \angle OPB$$ 6. From the figure, this is only possible if the ray $PO'$ coincides with the ray $PO$, since a part cannot equal the whole.
7. Hence, our assumption was incorrect, and the perpendicular to the tangent at the point of contact must pass through the centre $O$.
Hence proved.
1. Suppose, if possible, that the perpendicular to $AB$ at $P$ does not pass through the centre $O$.
2. Then let it pass through some other point $O'$. Join $O'P$.
3. By our assumption, $O'P \perp AB$: $$\angle O'PB = 90^\circ \quad \text{--- (1)}$$ 4. But we know from Theorem 1 that the line joining the centre $O$ to the point of contact $P$ is perpendicular to the tangent $AB$: $$OP \perp AB \implies \angle OPB = 90^\circ \quad \text{--- (2)}$$ 5. Comparing equations (1) and (2): $$\angle O'PB = \angle OPB$$ 6. From the figure, this is only possible if the ray $PO'$ coincides with the ray $PO$, since a part cannot equal the whole.
7. Hence, our assumption was incorrect, and the perpendicular to the tangent at the point of contact must pass through the centre $O$.
Hence proved.