3 Marks Question

Question 13 Marks

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.

Answer

We know that the tangent at any point of a circle is $\perp$ to the radius through the point of contact.
$ \therefore \angle O P A=90^{\circ} $
$ \therefore O A^2=O P^2+A P^2[B y \text { Pythagoras theorem }] $
$ \Rightarrow(5)^2=(O P)^2+(4)^2 $
$ \Rightarrow 25=(O P)^2+16 $
$ \Rightarrow O P^2=9 $
$ \Rightarrow O P=3 \mathrm{~cm}$

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Question 23 Marks

Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.

Answer

Let $APB$ be the tangent and take $O$ as centre of the circle.

Let us suppose that MP$\bot AB$ does not pass through the centre.
Then,
$\angle OPA = 90^\circ$ [$\because$ Tangent is perpendicular to the radius of circle]
But $\angle MPA = 90^\circ$ [Given]
$\therefore \angle OPA = \angle MPA$
This is only possible when point $O$ and point $M$ coincide with each other.
Hence, the perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.

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Question 33 Marks

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Answer

Given: $P Q$ is a diameter of a circle with centre $O$.
The lines $A B$ and $C D$ are the tangents at $P$ and $Q$ respectively.
To Prove: $A B \| C D$
Proof: Since $A B$ is a tangent to the circle at $P$ and $O P$ is the radius through the point of contact.
$\therefore \angle O P A=90^{\circ} \ldots \ldots(i)$
[The tangent at any point of a circle is $\perp$ to the radius through the point of contact]
$\because CD$ is a tangent to the circle at $Q$ and $OQ$ is the radius through the point of contact.
$\therefore \angle O Q D=90^{\circ} \ldots \ldots \ldots \text { (ii) }$
[The tangent at any point of a circle is $\perp$ to the radius through the point of contact]
From eq. $(i)$ and $(ii)$, $\angle O P A=\angle O Q D$
But these form a pair of equal alternate angles also,
$\therefore AB \| CD$

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Question 43 Marks

Prove that parallelogram circumscribing a circle is a rhombus.

Answer

Given $ABCD$ is a parallelogram in which all the sides touch a given circle
To prove:- $ABCD$ is a rhombus
Proof:-
$\because$ $ABCD$ is a parallelogram
$\therefore AB = DC$ and $AD = BC$
Again $AP, AQ$ are tangents to the circle from the point A
$\therefore AP = AQ$
Similarly, $BR = BQ$
$CR = CS$
$DP = DS$
$\therefore (AP + DP) + (BR + CR) = AQ + DS + BQ + CS = (AQ + BQ) + (CS + DS)$
$\Rightarrow AD + BC = AB + DC$
$\Rightarrow BC + BC = AB + AB [\because AB = DC, AD = BC]$
$\Rightarrow 2BC = 2AB$
$\Rightarrow BC = AB$
Hence, parallelogram $ABCD$ is a rhombus

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Question 53 Marks

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.

Answer

$\angle O A P=90^{\circ}....... (1)$ [Angle between tangent and radius through the point of contact is $90^{\circ}$ ]
$\angle O B P=90^{\circ} .......(2)$ [Angle between tangent and radius through the point of contact is $90^{\circ}$ ]
$\therefore OAPB$ is quadrilateral

$ \therefore \angle A P B+\angle A O B+\angle O A P+\angle O B P=360^{\circ}[\text { Angle sum property of a quadrilateral }] $
$ \Rightarrow \angle A P B+\angle A O B+90^{\circ}+90^{\circ}=360^{\circ}[\text { From (1) and (2) }] $
$ \Rightarrow \angle A P B+\angle A O B=180^{\circ} $
$\Rightarrow \angle A P B \text { and } \angle A O B \text { are supplementary }$

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Question 63 Marks

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

Answer

Steps of Construction:
i. Draw a circle with any radius and center $O$. Here $x y$ is given line.
ii. Choose any point $P$ on the circumference of the circle, and draw a line passing through $P$, Let's name it $A B$.
iii. Draw a line $A B$ parallel to $x y$, such that $A B$ intersects the circle at two points $P$ and $A$.Here, $A B$ and $x y$ are two parallel lines. $A B$ intersects the circle at exactly two points, $P$ and $Q$. Therefore, line $A B$ is the secant of this circle.
iv. $C D$ intersects the circle at exactly one point, $R$, line $C D$ is the tangent to the circle.

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Question 73 Marks

Two tangents $TP$ and $TQ$ are drawn to a circle with centre $O$ from an external point $T$. Prove that $\angle PTQ = 2 \angle OPQ.$

Answer

Given $A$ circle with centre $O$ and an external point $T$ and two tangents $T P$ and $T Q$ to the circle, where $P, Q$ are the points of contact.
To Prove: $\angle PTQ = 2 \angle OPQ$
Proof: Let $\angle PTQ = \theta$
Since $TP, TQ$ are tangents drawn from point $T$ to the circle.
$TP = TQ$
$\therefore$ TPQ is an isoscles triangle
$\therefore \angle \mathrm{TPQ}=\angle \mathrm{TQP}=\frac{1}{2}\left(180^{\circ}-\theta\right)=90^{\circ}-\frac{\theta}{2}$
$\text { Since, TP is a tangent to the circle at point of contact } P$
$ \therefore \angle \mathrm{OPT}=90^{\circ} $
$ \therefore \angle \mathrm{OPQ}=\angle \mathrm{OPT}-\angle \mathrm{TPQ}=90^{\circ}-\left(90^{\circ}-\frac{1}{2} \theta\right)=\frac{\theta}{2}=\frac{1}{2} \angle \mathrm{PTQ}$
Thus, $\angle PTQ = 2 \angle OPQ$

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Question 83 Marks

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 (see figure). Find the sides AB and AC.

Answer

Join OE and OF. Also join OA, OB and OC.

Since BD = 8 cm
$\therefore$ BE = 8 cm
[Tangents from an external point to a circle are equal]
Since CD = 6 cm
$\therefore$ CF = 6 cm
[Tangents from an external point to a circle are equal]
Let AE = AF = x
Since OD = OE = OF = 4 cm [Radii of a circle are equal]
$\therefore$ Semi-perimeter of $\triangle$ ABC = $\frac{(x\;+6)(x\;+8)\;+\;(6+8)}2 = \frac{(2x\;+28)}2$= (x + 14)cm
$\therefore$ Area of $\triangle$ABC =$\sqrt{s(s-a)(s-b)\;(s-c)}$
= $\sqrt{(x\;+14)(x\;+14\;-14)(x\;+14-\overline{x+8})(x\;+14-\overline{x+6})}$
= $\sqrt{(x\;+14)(x\;)(8)(6)}$cm²
Now, Area of ΔABC = Area of $\triangle$OBC + Area of $\triangle$OCA + Area of $\triangle$OAB
$\Rightarrow$ $\sqrt{(x\;+14)(x\;)(8)(6)}$=$\frac{(6+8)4}2+\frac{\displaystyle(x+6)4}{\displaystyle2}+\frac{\displaystyle(x+8)4}{\displaystyle2}$
$\Rightarrow$ $\sqrt{(x\;+14)(x\;)(8)(6)}$= 28 + 2x + 12 + 2x + 16
$\Rightarrow$ $\sqrt{(x\;+14)(x\;)(8)(6)}$= 4x + 56
$\Rightarrow$$\sqrt{(x\;+14)(x\;)(8)(6)}$=4(x + 14)
Squaring both sides,
(x + 14) (x) (6) (8) = 16(x + 14)²
$\Rightarrow$ 3x = x + 14
$\Rightarrow$ 2x = 14
$\Rightarrow$ x = 7
$\therefore$ AB = x + 8 = 7 + 8 = 15 cm
And AC = x + 6= 7 + 6 = 13 cm

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Question 93 Marks

Prove that the lengths of tangents drawn from an external point to a circle are equal.

Answer

Consider a circle with center O and an external point P. Let PA and PB be two tangents drawn from point P to the circle, where A and B are the points of tangency. We need to prove that PA = PB.
To prove this, we can construct two right-angled triangles by joining the center O to the points of tangency A and B, and to the external point P. This gives us $\triangle O A P$ and $\triangle O B P$.
Here are the reasons why the two triangles are congruent:
OA = OB (Radii of the same circle).
OP = OP (Common side to both triangles).
$\angle O A P=\angle O B P=90^{\circ}$ (The tangent at any point on a circle is perpendicular to the radius through the point of contact).
Based on these three points, we can conclude that $\triangle O A P \cong \triangle O B P$ by the RHS (Right Angle-Hypotenuse-Side) congruence criterion.
Since the two triangles are congruent, their corresponding parts must be equal. Therefore PA = PB (by CPCT - Corresponding Parts of Congruent Triangles).
Hence, the lengths of the tangents drawn from an external point to a circle are equal.

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