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

Solution :

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