NCERT Solutions: Class 10 Maths Chapter 10 - Circles

At every point of a circle there is exactly one tangent. Since a circle has infinitely many points on it, it can have infinitely many tangents.

A tangent touches a circle at exactly one point. A line meeting a circle at two points is a secant. For any direction, at most two parallel tangents can touch the circle, one on each side. The point where a tangent touches the circle is the point of contact.

1. A. 12 cm
2. B. 13 cm
3. C. 8.5 cm
4. D. $\sqrt{119}$ cm

The radius to the point of contact is perpendicular to the tangent, so $OP \perp PQ$. In right triangle $OPQ$, $OP=5$ cm and $OQ=12$ cm. Thus $PQ=\sqrt{OQ^2-OP^2}=\sqrt{12^2-5^2}=\sqrt{119}$ cm.

Draw a circle with centre $O$. Choose a line $l$ whose perpendicular distance from $O$ is less than the radius; it will cut the circle in two points, so it is a secant. Now draw a line $m$ parallel to $l$ whose perpendicular distance from $O$ is exactly the radius; it touches the circle at one point, so it is a tangent.

1. A. 7 cm
2. B. 12 cm
3. C. 15 cm
4. D. 24.5 cm

Let $O$ be the centre and $P$ the point of contact. Since $OP \perp QP$, triangle $OQP$ is right-angled at $P$. Therefore $OP=\sqrt{OQ^2-QP^2}=\sqrt{25^2-24^2}=\sqrt{49}=7$ cm.

1. A. $60^\circ$
2. B. $70^\circ$
3. C. $80^\circ$
4. D. $90^\circ$

Radii to the points of contact are perpendicular to the tangents, so $\angle OPT=90^\circ$ and $\angle OQT=90^\circ$. In quadrilateral $OPTQ$, $\angle PTQ=360^\circ-90^\circ-90^\circ-110^\circ=70^\circ$.

1. A. $50^\circ$
2. B. $60^\circ$
3. C. $70^\circ$
4. D. $80^\circ$

The line joining the centre to the external point bisects the angle between the two tangents, so $\angle OPA=40^\circ$. Also $OA \perp PA$, hence $\angle OAP=90^\circ$. In triangle $OPA$, $\angle POA=180^\circ-90^\circ-40^\circ=50^\circ$.

Let $AB$ be a diameter of a circle with centre $O$. Let tangents at $A$ and $B$ be $l$ and $m$. Since a tangent is perpendicular to the radius through the point of contact, $l \perp OA$ and $m \perp OB$. But $OA$ and $OB$ lie on the same straight line $AB$. Hence both $l$ and $m$ are perpendicular to the same line $AB$, so $l \parallel m$.

Let a tangent touch the circle at $P$, and let $O$ be the centre. By the tangent-radius theorem, $OP$ is perpendicular to the tangent at $P$. Through a given point on a line, only one perpendicular can be drawn to that line. Therefore the perpendicular at $P$ to the tangent must be the line $OP$, and hence it passes through the centre $O$.

Let $O$ be the centre and $P$ the point of contact. Then $OP \perp AP$, $OA=5$ cm and $AP=4$ cm. By Pythagoras theorem, $OP=\sqrt{OA^2-AP^2}=\sqrt{5^2-4^2}=\sqrt9=3$ cm.

Let $AB$ be the chord of the larger circle touching the smaller circle at $P$, and let $O$ be the common centre. Since $OP$ is perpendicular to the tangent chord $AB$, it bisects $AB$. Here $OA=5$ cm and $OP=3$ cm. Thus $AP=\sqrt{OA^2-OP^2}=\sqrt{25-9}=4$ cm, so $AB=2AP=8$ cm.

Let the circle touch $AB, BC, CD, DA$ at $P, Q, R, S$ respectively. Tangents from the same external point are equal, so $AP=AS$, $BP=BQ$, $CQ=CR$ and $DR=DS$. Now $AB+CD=(AP+BP)+(CR+DR)$ and $AD+BC=(AS+DS)+(BQ+CQ)$. Substituting the equal tangent lengths gives $AB+CD=AD+BC$.

Let the upper and lower parallel tangents touch the circle at $P$ and $Q$. From the external point $A$, the line $AO$ bisects the angle between the two tangents $AP$ and $AC$. From the external point $B$, the line $BO$ bisects the angle between the two tangents $BQ$ and $BC$. Since $AP \parallel BQ$, the interior angles made with the transversal $AB$ are supplementary. Therefore their halves add to $90^\circ$, so the angle between $AO$ and $BO$ is $90^\circ$. Hence $\angle AOB=90^\circ$.

Let tangents $PA$ and $PB$ touch a circle with centre $O$ at $A$ and $B$. Since $OA \perp PA$ and $OB \perp PB$, we have $\angle OAP=90^\circ$ and $\angle OBP=90^\circ$. In quadrilateral $AOBP$, the angle sum is $360^\circ$, so $\angle APB+\angle AOB=360^\circ-180^\circ=180^\circ$. Hence the two angles are supplementary.

Let parallelogram $ABCD$ circumscribe a circle. For any tangential quadrilateral, $AB+CD=AD+BC$. In a parallelogram, opposite sides are equal, so $AB=CD$ and $AD=BC$. Therefore $2AB=2AD$, giving $AB=AD$. Hence all four sides are equal, and the parallelogram is a rhombus.