3 Marks Question

Question 13 Marks

Out of the two concentric circles, the radius of the outer circle is 5cm and the chord $AC$ of length 8cm is a tangent to the inner circle. Find the radius of the inner circle.

Answer

Let $C_1$ and $C_2$ be the two circle having same centre $O$. $A C$ is a chord which touches the $C_1$ at point $D$. Join OD.
Also, $O D \perp A C$
$\therefore A D=D C=4 cm$ [perpendicular line $O D$ bisects the chord]
In right angled $\triangle\text{AOD},\ \ \text{OA}^2=\text{AD}^2+\text{DO}^2$
[by Pythagoras theorem, i.e., (hypotenuse) $\left.{ }^2=(\text { base })^2+(\text { perpendicular) })^2\right]$
$\Rightarrow DO^2 = 5^2 - 4^2$
$\Rightarrow 25 - 16 = 9$
$\Rightarrow DO = 3cm$
$\therefore$ Radius of the inner circle $OD = 3cm.$

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Question 23 Marks

If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that $\angle\text{DBC}=120^\circ,$ prove that BC + BD = BO, i.e., BO = 2BC.

Answer

Two tangents BD and BC are drawn from an external point B.

To prove BO = 2BC
Given, $\angle\text{OBC}=120^\circ$
Join OC, OD and BO.
Since, BC and BD are tangents.
$\therefore\ \text{OC}\perp\text{BC and OD}\perp\text{BD}$
we know, OB is a angle bisector of $\angle\text{DBC}.$
$\therefore\ \angle\text{OBC}=\angle\text{DBO}=60^\circ$
In right angled $\triangle\text{OBC},\ \ \cos60^\circ=\frac{\text{BC}}{\text{OB}}$
$\Rightarrow\ \frac{1}{2}=\frac{\text{BC}}{\text{OB}}$
⇒ OB = 2BC
Also, BC = BD
[tangent drawn from internal point to circle are equal]
$\therefore$ OB = BC + BD
⇒ OB = BC + BD

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Question 33 Marks

Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A.

Answer

Given, AB is a diameter of the circle. A tangent is drawn from point A. Draw a chord CD parallel to the tangent MAN.

So, CD is a chord of the circle and OA is a radius of the circle. $\angle\text{MAO}=90^\circ$ [tangent at any point of a circle is perpendicular to the radius throgh the point of contact] $\angle\text{CEO}=\angle\text{MAO}$ [corresponding angles] $\therefore\ \angle\text{CEO}=90^\circ$ Thus, OE bisects CD, [perpendicular from centre of circle to chord bisects the chord] Similarly, the diameter AB bisects all. Chords which are parallel to the tangent at the point A.

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Question 43 Marks

A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ.

Answer

Given chord PQ is parallel to tangent at R. To prove R bisects the are PRQ

Proof: $\angle1=\angle2$ [alternate interior angles] $\angle1=\angle3$ [angle between tangent and chord is equal to angle made by chord in alternate segment] $\therefore\ \ \angle2=\angle3$ ⇒ PR = QR [sides opposite to equal angles are equal] ⇒ PR = QR So, R bisects PQ.

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Question 53 Marks

In Figure, common tangents AB and CD to two circles intersect at E. Prove that AB = CD.

Answer

Given common tangent AB and CD to two circles intersecting at E. To prove AB = CD

Proof: EA = EC ...(i) [the lengths of tangents drawn from an internal point to a circle are equal] EB = ED ...(ii) On adding Eqs. (i) and (ii), we get EA + EB = EC + ED ⇒ AB = CD Hence proved.

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