# Circles

## Class 10 NCERT

### NCERT

1   $[ \frac{-b+\sqrt{-b^2-4ac}}{2a}]$

##### Solution :

2   How many tangents can a circle have?

##### Solution :

A circle can have an infinite number of tangents.

3   Fill in the blanks : $\\$ i) A tangent to a circle intersects it in _________ point(s).$\\$ (ii) A line intersecting a circle in two points is called a _______.$\\$ (iii) A circle can have _______ parallel tangents at the most.$\\$ (iv) The common point of a tangent to a circle and the circle is called ______.

##### Solution :

i) A tangent to a circle intersects it in exactly one point(s).$\\$ (ii) A line intersecting a circle in two points is called a secant.$\\$ (iii) A circle can have two parallel tangents at the most.$\\$ (iv) The common point of a tangent to a circle and the circle is called Point of Contact.

4   A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length of PQ is : $\\$ A) 12 cm (B) 13 cm (C) 8.5 cm (D) 119

##### Solution :

(D) is the answer.$\\$ Because, PQ = $\sqrt{(OQ^2-OP^2)}$ = $\sqrt{(12^2-5^2)}$ = $\sqrt{119}$

5   Draw a circle and two lines parallel to a given line such that one is tangent and the other, a secant to the circle.

##### Solution :

From the Given Figure below,

Let $'I'$ be the given line and a circle with given $o$ is drawn.

$\bullet$ Line PT is drawn $\|$ to line 'I'

$\bullet$ PT is the tangent to the circle

$\bullet$ AB is drawn $\|$ to line 'I' and is the secant

6   From a point $Q$, the length of the tangent to a circle is $24 cm$ and the distance of $Q$ from the center is $25 cm$. The radius of the circle is

Let $‘O’$ be the centre of the circle Given: $\\$$\bullet Distance of Q from the centre, OQ = 25cm \\$$ \bullet$ Length of the tangent to a circle, $PQ = 24 cm$ $\\$$\bullet Radius, OP = ? We know that, Radius is perpendicular to the tangent at the point of contact Hence, OP \perp PQ Therefore, OPQ forms a Right Angled Triangle Applying Pythagoras theorem for \Delta OPQ, OP^2+ PQ^2 = OQ^2 By substituting the values in the above Equation, OP^2 + 24^2 = 25^2 OP^2 = 625 – 576 \left(By Transposing\right) OP^2 = 49 OP = 7 \left(By Taking Square Root \right) Therefore, the radius of the circle is 7 cm. Hence, alternative\left (A \right) is correct. 7 In the given figure, if TP and TQ are the two tangents to a circle with centre O so that \angle POQ = 110^°, then \angle PTQ is equal to ##### Solution : Given: \\$$\bullet$Tangents: $TP$ and $TQ$ We know that, Radius is perpendicular to the tangent at the point of contact Thus, $OP \perp TP$ and $OQ \perp TQ$ $\\$$\bullet Since the Tangents are Perpendicular to Radius \circ \angle OPT = 90^o \\ \circ \angle OQT = 90^o \\ Now, POQT forms a Quadrilateral \\ We know that, Sum of all interior angles of a Quadrilateral = 360^o \\ \implies 90^o + 110^o + 90^o + PTQ = 360^o \left(By Substituting \right) \\ \implies \angle PTQ = 70^o \\ Hence, alternative \left(B\right) is correct. 8 If tangents PA and PB from a point P to a circle with centre O are inclined to each other an angle of 80^o, then \angle POA is equal to ##### Solution : \angle PaO = \dfrac{1}{2} \angle AOB = \dfrac{100^o}{2} = 50^o Hence, alternative \left(A\right) is correct. \angle OAP + \angle APB +\angle PBO + \angle BOA = 360^o 90^o + 80^o +90^o + \angle BOA = 360^o \left(By Substituting \right) A \leftrightarrow B, P \leftrightarrow P, O \leftrightarrow O And thus, \angle POB = \angle POA \\ \angle PaO = \dfrac{1}{2} \angle AOB = \dfrac{100^o}{2} = 50^o Hence, alternative \left(A\right) is correct. Now, AOBP forms a Quadrilateral We know that, \\ Sum of all interior angles of a Quadrilateral = 360^o \angle BOA = \angle AOB = 100^o \\ In \Delta OPB and \Delta OPA, \\ AP = BP \left(Tangents from a point \right) \\ OA = OB \left(Radii of the circle \right) \\ OP = OP \left(Common side \right) \\ Therefore, \Delta OPB \cong \Delta OPA \left(SSS congruence criterion \right) \\ A \leftrightarrow B, P \leftrightarrow P, O \leftrightarrow O And thus, \angle POB = \angle POA \\ \angle PaO = \dfrac{1}{2} \angle AOB = \dfrac{100^o}{2} = 50^o Hence, alternative \left(A\right) is correct. Given: \\$$\bullet$ Tangents are $PA$ and $PB$ $\\$We know that, Radius is perpendicular to the tangent at the point of contact Thus, $OA \perp PA$ and $OB \perp PB$ $\\$$\bullet Since the Tangents are Perpendicular to Radius\\ \circ \angle OBP = 90^o \\ \circ \angle OAP = 90^o \\ Now, AOBP forms a Quadrilateral We know that, \\ Sum of all interior angles of a Quadrilateral = 360^o \\ \angle OAP + \angle APB +\angle PBO + \angle BOA = 360^o 90^o + 80^o +90^o + \angle BOA = 360^o \left(By Substituting \right) \\ \angle BOA = \angle AOB = 100^o \\ In \Delta OPB and \Delta OPA, \\ AP = BP \left(Tangents from a point \right) \\ OA = OB \left(Radii of the circle \right) \\ OP = OP \left(Common side \right) \\ Therefore, \Delta OPB \cong \Delta OPA \left(SSS congruence criterion \right) \\ A \leftrightarrow B, P \leftrightarrow P, O \leftrightarrow O And thus, \angle POB = \angle POA \\ \angle PaO = \dfrac{1}{2} \angle AOB = \dfrac{100^o}{2} = 50^o Hence, alternative \left(A\right) is correct. 9 Prove that the tangents drawn at the ends of a diameter of a circle are parallel. ##### Solution : From the figure, Given \\ \bullet Let PQ be a diameter of the circle. \\ \bullet Two tangents AB and CD are drawn at points P and Q respectively. To Prove\colon Tangents drawn at the ends of a diameter of a circle are parallel. Proof\colon We know that, Radius is perpendicular to the tangent at the point of contact Thus, OP \perp AB and OQ \perp CD Since the Tangents are Perpendicular to Radius \circ \angle OQC = 90^o \\ \circ \angle OQD = 90^o \\ \circ \angle OPA = 90^o \\ \circ \angle OPB = 90^o \\ From Observation, \circ \angle OPC = \angle OQB \left(Alternate interior angles\right) \\ \circ \angle OPD = \angle OQA \left(Alternate interior angles \right) If the Alternate interior angles are equal then lines AB and CD should be parallel. We know that AB \& CD are the tangents to the circle. Hence, it is proved that Tangents drawn at the ends of a diameter of a circle are parallel. 10 Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre. ##### Solution : From the figure, \\ Given \colon \\ \circ Let ‘O’ be the centre of the circle \\$$\circ$ Let $AB$ be a tangent which touches the circle at $P$. To Prove$\colon$ $\circ$ Line perpendicular to $AB$ at $P$ passes through centre $O$. $\\$ Proof$\colon$ Consider the figure below,

Let us assume that the perpendicular to $AB$ at $P$ does not pass through centre $O$. $\\$ Let It pass through another point $Q$. Join $OP$ and $QP$. $\\$ We know that, $\\ \ \ \$Radius is perpendicular to the tangent at the point of contact Hence,

$AB \perp PQ$

$\therefore \angle QPB = 90^o \dots \left(1\right)$

We know the line joining the centre and the point of contact to the tangent of the circle are perpendicular to each other.$\\$ $\therefore \angle QPB = 90^o \dots \left(2\right)$

Comparing equations $\left(1\right)$ and $\left(2\right)$,$\\$we obtain $\therefore \angle QPB = \angle OPB \dots \left(3\right)$ $\\$From the figure, it can be observed that,$\\$$\therefore \angle QPB < \angle OPB \dots \left(4\right) Therefore, in reality \angle QPB \neq \angle OPB \\ \angle QPB = \angle OPB only if QP = OP which is possible in a scenario when the line QP coincides with OP. Hence it is proved that the perpendicular to AB through P passes through centre O. 11 The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle. ##### Solution : From the Figure\colon$$\\$

Given$\colon$ $\circ$ Let point $‘O’$ be the centre of a circle$\\$ $\circ AB$ is a tangent drawn on this circle from point $A, AB = 4 cm$ $\\$ $\circ$Distance of $A$ from the centre, $OA = 5cm$ $\\$ $\circ$ Radius, $OB = ?$

In $\Delta ABO$, We know that,$\\$ $OB \perp AB$ $\left(\text {Radius$\perp$tangent at the point of contact}\right)$

$OAB$ forms a Right Angled Triangle. Hence using, Pythagoras theorem

in $\angle ABO,$ $\\$ $AB^2 + OB^2 = OA^2$ $\\$ $42 + OB^2 = 5^2$ (By Substituting) $\\$ $16 + OB^2 = 25$ $\\$ $OB^2 = 9$ $\\$ Radius, $OB = 3$ (By Taking Square Roots)$\\$ Hence, the radius of the circle is $3 cm$.

12   Two concentric circles are of radii $5 cm$ and $3 cm$ . Find the length of the chord of the larger circle which touches the smaller circle.

##### Solution :

From the Figure,

Given, $\\$ $\circ$ Let $‘O’$ be the centre of the two concentric circles $\\$$\circ Let PQ be the chord of the larger circle which touches the smaller circle at point A. \\$$\circ$ $PQ = ?$

By Observation, $\\$ Line $PQ$ is tangent to the smaller circle. $\\$Hence, $OA \perp PQ \left(Radius \perp tangent at the point of contact\right)$$\\$$ \Delta OAP$ forms a Right Angled Triangle By applying Pythagoras theorem in $\Delta OAP$

$OA^2 + AP^2 = OP^2 \\ 32 + AP^2 = 52 \left(By Substituting\right) \\ 9 + AP^2 = 25 \\ AP^2 = 16 \\ AP = 4 \left(By Taking Square Roots\right)$

In $\Delta OPQ$ ,$\\$ Since $OA \perp PQ, AP = AQ \left(Perpendicular from the center of the circle bisects the chord\right)$ $\\$ $\therefore PQ = 2 times AP = 2 × 4 = 8 \left(Substituting AP = 4cm \right)$ $\\$Therefore, the length of the chord of the larger circle is $8 cm.$

13   A quadrilateral $ABCD$ is drawn to circumscribe a circle $\left(see given figure\right)$ Prove that$AB + CD = AD + BC$

##### Solution :

From the Figure,$\\$ Given,$\\$ $\circ DC , DA, BC, AB$ are sides of the Quadrilaterals which also form the tangents to the circle inscribed within Quadrilateral $ABCD$$\\ To Prove\colon$$\\$ $AB + CD = AD + BC$

Proof$\colon$ We know that length of tangents drawn from an external point of the circle are equal.$\\$ $DR = DS \dots \left(1\right)$$\\ CR = CQ \dots \left(2\right)$$\\$ $BP = BQ \dots \left(3\right)$$\\ AP = AS \dots \left(4\right) Adding \left(1\right), \left(2\right), \left(3\right), \left(4\right),\\ we obtain DR + CR + BP + AP = DS + CQ + BQ + AS \left(DR + CR\right) + \left(BP + AP\right) \\ = \left(DS + AS\right) + \left(CQ + BQ\right) \left(By regrouping\right) \dots \left(5\right) From the figure,\\ \circ DR +CR = DC \\ \circ BP + AP = AB \\ \circ DS + AS = AD \\ \circ CQ +BQ = BC$$\\$ Hence substituting the above values in Equation $\left(5\right)$$\\ CD + AB = AD + BC Hence it is proved. 14 In the given figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that \Delta AOB = 90^o. ##### Solution : From the Figure,\\Given,\\ \bullet Let ‘O’ be the centre of the circle \\ \bullet XY and X’Y’ are two parallel tangents to circle\\ \bullet AB is another tangent such that with point of contact C intersecting XY at A and X’Y’ at B. To Prove\colon \angle AOB=90^o. Proof \colon \\ Join point O to C. From the Figure above, Consider \\$$\Delta OPA$ and $\Delta OCA,$

Here, $\\$ $\circ OP = OC \left(Radii of the same circle \right)$ $\\$ $\circ AP = AC \left(Tangents from external point A \right)$$\\ \circ AO = AO \left(Common side\right) Therefore, \Delta OPA \cong \Delta OCA \left(SSS congruence criterion\right) Hence,\\$$ P \leftrightarrow C, A \leftrightarrow A, O \leftrightarrow O$ We can also say that,

$\angle POA = \angle COA \dots \left(i\right)$$\\ Similarly, \Delta OQB \cong \Delta OCB$$\\$ $\angle QOB = \angle COB \dots \left(ii\right)$$\\ Since POQ is a diameter of the circle, it is a straight line. \\Therefore, \angle POA + \angle COA + \angle COB + \angle QOB = 180^o \dots \left(3\right) Substituting Equation \left(i\right) and \left(ii\right) in the Equation \left(3\right), \\ 2\angle COA + 2 \angle COB = 180^o \\ 2\left(\angle COA + \angle COB\right) = 180^o \\ \angle COA + \angle COB = 90^o \left(By Transposing\right) \angle AOB = 90^o 15 Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre. ##### Solution : From the figure,\\ Given,\\ \circ Let us consider a circle centered at point O.\\ \circ Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively \\ \circ AB is the line segment, joining point of contacts A and B together such that it subtends \angle AOB at center O of the circle. \\ To Prove \colon$$\\$ $\angle APB$ is supplementary to $\angle AOB$

Proof$\colon$ Join $OP$ $\\$

Consider the $\Delta OAP \& \Delta OBP,$ $\\$ $\circ PA = PB\left( Tangents drawn from an external point are equal\right)$$\\ \circ OA = OB \left( Radii of the same circle\right) \circ OP = OP\left(Common Side\right)$$\\$ Therefore, $\Delta OAP \cong \Delta OBP \left(SSS congruence criterion\right)$ Hence, $\\$ $\circ \angle OPA = \angle OPB$$\\ \circ \angle AOP = \angle BOP Also,\\ \circ \angle APB = 2 \angle OPA \dots \left(1\right)$$\\$ $\circ \angle AOB= 2 \angle AO \dots \left(2\right)$

In the Right angled Triangle $\Delta OAP,$ $\angle AOP +\angle OPA = 90^o \\ \angle AOP = 90^o - \angle OPA \dots \left(3\right)$$\\ Multiplying the Equation \left(3\right) by 2, \\ 2\angle AOP = 180^o - 2\angle OPA$$\\$ By Substituting $\left(1\right)$ and $\left(2\right)$ in the equation above,$\\$ $\angle AOB = 180^o - \angle APB \\ \angle AOP + \angle OPA = 180^o $$\\Hence it is provehat the angd tle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre. 16 Prove that the parallelogram circumscribing a circle is a rhombus. ##### Solution : From the figure, \\ Given, \\ \circ ABCD is a parallelogram, \\ \circ Hence \\ \ \ \ \ \circ AB = CD \dots \left(\right)\\ \ \ \ \ \circ BC = AD \dots \left(2\right) \\ To Prove\colon$$\\$ $\circ$Parallelogram circumscribing a circle is a rhombus.

Proof$\colon$ $\\$ From the figure, $\\$ $\circ DR = DS \left(Tangents on the circle from point D\right) \\ \circ CR = CQ \left(Tangents on the circle from point C\right) \\ \circ BP = BQ \left(Tangents on the circle from point B\right) \\ \circ AP = AS \left(Tangents on the circle from point \right)$

Adding all the above equations, we obtain$\\$ $DR + CR + BP + AP = DS + CQ + BQ + AS \left(DR + CR\right) + \left(BP + AP\right) = \left(DS + AS\right) + \left(CQ + BQ\right) \left(By Rearranging\right) \left(3\right)$ $\\$ From the figure,$\\$ $\circ DR + CR = CD, \\ \circ \left(BP + AP\right) = AB \\ \circ \left(DS + AS\right) = AD\\ \circ \left(CQ + BQ\right) = BC$

Substituting the above values in $\left(3\right),\\ CD + AB = AD + BC \dots \left(4\right)$ $\\$ On putting the values of equations $\left(1\right)$ and $\left(2\right)$ in the equation $\left(4\right)$, we obtain$\\$ $2AB = 2BC \\ AB = BC \dots \left(5\right)$ $\\$ Comparing equations $\left(1\right), \left(2\right), and \left(5\right)$, we get$\\$ $AB = BC = CD = DA$satisfies the property of Rhombus.$\\$ Hence, $ABCD$ is a rhombus.

17   A triangle $ABC$ is drawn to circumscribe a circle of radius $4 cm$ such that the Segments $BD$ and $DC$ into which $BC$ is divided by the point of contact $D$ are of lengths $8 cm$ and $6 cm$ respectively $\left(see given figure\right)$. Find the sides $AB$ and $AC$.

##### Solution :

From the figure, $\\$

Given, $\\$ $\bullet$ Let the given circle touch the sides $AB$ and $AC$ of the triangle at point $E$ and $F$ respectively $\\$ $\bullet$Length of the line segment $AF$ be $x$.$\\$ $\bullet In \Delta ABC,$

$\ \ \ \ \ \circ CF = CD = 6 cm \left(Tangents on the circle from point C\right) \\ \ \ \ \ \ \circ BE = BD = 8cm \left(Tangents on the circle from point B\right) \\ \ \ \ \ \ \circ AE = AF = x \left(Tangents on the circle from point A\right) \\ \ \ \ \ \ \circ AB = ? \\ \ \ \ \ \ \circ AC = ?$

$In \Delta ABE, AB = AE + BE = x + 8 \\ BC = BD + DC = 8 + 6 = 14\\ CA = CF + AF = 6 + x $$\\ We know that, \\ 2s = AB + BC + CA \\ = x + 8 + 14 + 6 + x \\ = 28 + 2x \\ s = 14 + x$$\\$ We also known that,$\\$

Area of $\Delta ABC = \sqrt{s\left( s - a \right)\left( s - b \right)\left( s - c \right)}$

$=\sqrt{\{14+x\} \{(14+x)-(6 + x ) \}\{ ( 14 + x )-( 8 +x )\} }$ $\\$ $=\sqrt{(14+x)(x)(8)(6)}$ $\\$ $=4 \sqrt{3(14x+x^2)}$ $\\$

Area of $\Delta OBC = \dfrac{1}{2} × OD × BC = \dfrac{1}{ 2} × 4 × 14 = 28$

Area of $\Delta OCA = \dfrac{1}{2} × OF × AC = \dfrac{1}{ 2} × 4 × \left( 6 + x \right) = 12 + 2x$

Area of $\Delta OAB = \dfrac{1}{ 2} × OE × AB = \dfrac{1}{ 2} × 4 × \left( 8 + x \right) = 16 + 2x$

Area of $\Delta ABC = Area of \Delta OBC + Area of \Delta OCA + Area of \Delta OAB \\ 4\sqrt{ 3\left(14x + x^2 \right)} = 28 + 12 + 2x + 16x + 2x \\ \implies 4 \sqrt{3\left(14x + x^2 \right)} = 56 + 4x$

$\implies \sqrt{3 \left( 14x + x^2 \right)} = 14 + x \\ \implies 3 \left( 14x + x 2 \right) = \left( 14 + x^2 \right)$

$\implies 42x + 3x^2 = 196 + x^2 + 28x\\ \implies 2x^2 + 14x - 196 = 0 \\ \implies x^2+ 7x - 98 = 0 \\ \implies x^2+ 14x - 7x - 98 = 0$

$\implies \left( x + 14 \right)-7\left( x +14 \right) = 0$ $\\$ $\implies \left( x + 14 \right)\left( x - 7 \right) = 0$

Either $x +14 = 0$

or $x-7 = 0$

Therefore, $x = -14$ and $7$ $\\$ However, $x = -14$ is not possible as the length of the sides will be negative. $\\$ Therefore, $x = 7$ $\\$ Hence, $AB = x + 8 = 7 + 8 = 15 cm$ $\\$ $CA = 6 + x = 6 + 7 = 13 cm$

18   Prove that opposite sides of a quadrilateral circumscribing a circle subtend Supplementary angles at the centre of the circle.

##### Solution :

From the figure ,$\\$

Given,$\\$ To Prove$\colon $$\\ \circ Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.\\ \circ i.e., \angle AOB + COD = 180^o \& \angle BOC + \angle DOA = 180^o$$\\$ Proof$\colon$ $\\$ $\circ$ Let us join the vertices of the quadrilateral $ABCD$ to the center of the circle.$\\$ Consider $\Delta OAP$ and $\Delta OAS,$$\\ AP = AS \left(Tangents from the same point\right) \\ OP = OS \left(Radii of the same circle\right) \\ OA = OA \left(Common side\right)$$\\$ $\Delta OAP \cong \Delta OAS$(SSS congruence criterion) $\\$ Therefore, $A\leftrightarrow A, P \leftrightarrow S, O \leftrightarrow O$ $\\$ And thus, $\angle POA = \angle AOS\\ \angle 1 = \angle 8$ Similarly, $\angle 2 = \angle 3 \\ \angle 4 = \angle 5\\ \angle 6 = \angle 7 \\ \angle 1+ \angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8 = 360^o \\ \left (\angle 1 + \angle 8\right) + \left(\angle 2 + \angle 3\right ) + \left(\angle 4 + \angle 5\right) + \left(\angle 6 + \angle 7\right) = 360^o (By Rearranging) \\ 2\angle 1 + 2\angle 2 + 2\angle 5 + 2\angle 6 = 360^o \\ 2\left(\angle 1 +\angle 2\right) + 2\left(\angle 5 + \angle 6\right ) = 360^o \\ \left(\angle 1 +\angle 2\right) + 2\left(\angle 5 + \angle 6\right) = 180^o \\ \angle AOB +\angle COD = 180^o$ Similarly, we can prove that $\angle BOC + \angle DOA = 180^o$ Hence, opposite sides of a quadrilateral circumscribing a circle subtend Supplementary angles at the centre of the circle