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Question 11 Mark
A quadrilateral $ABCD$ is drawn to circumscribe a circle. Prove that $AB + CD = AD + BC$
Answer
We know that the lengths of tangents drawn from an exterior point to a circle are equal.
$AP = AS, ... (i)$ [tangents from $A$]
$BP = BQ, ... (ii)$ [tangents from $B$]
$CR = CQ, ... (iii)$ [tangents from $C$]
$DR = DS. ... (iv)$ [tangents from $D$]
$AB + CD = (AP + BP) + (CR + DR)$
$= (AS + BQ) + (CQ + DS) [using (i), (ii), (iii), (iv)]$
$= (AS + DS) + (BQ + CQ)$
$= AD + BC.$
Hence, $AB + CD = AD + BC.$
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Question 21 Mark
If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at angle of $80^\circ$, $\angle$$POA$ is equal to
Answer
Here $\angle $$APB = 80^\circ$
$\therefore $$\angle $$AOB = 180^\circ - 80^\circ = 180^\circ$
Now, since $OP$ bisect $\angle $$APB$ and $\angle $$AOB.$
$\therefore $$\angle $AOP = $\frac{{{{100}^ \circ }}}{2} = {50^ \circ }$
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Question 31 Mark
In Figure, if $TP$ and $TQ$ are two tangents to a circle with centre $O$ so that $\angle $$POQ =$$ {110^o}$, then $\angle$$PTQ$ is equal to:
Answer
self-learning
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Question 41 Mark
From a point $Q,$ the length of tangent to a circle is $24\ cm$ and the distance of $Q$ from the centre is $25\ cm$. The radius of the circle is
Answer

Here $\angle $$OPQ = 90^\circ [$Tangent makes right angle with the radius at the point of contact$]$
in right angled triangle $OPQ$
$ \therefore O Q^2=O P^2+P Q^2 \Rightarrow(25)^2=O P^2+(24)^2$
$ \Rightarrow O P^2=625-576$
$ \Rightarrow $$OP = 7\ cm$ Therefore, the radius of the circle is $7\ cm$
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Question 51 Mark
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 $PQ$ is:
Answer
We know that the line drawn from the centre of the circle to the tangent is perpendicular to the tangent.
$OP \perp PQ$
By applying Pythagoras theorem in $\triangle OPQ$,

$ O P^2+P Q^2=O Q^2 $
$ 5^2+P Q^2=12^2$
$ P Q^2=144-25$
$\mathrm{PQ}=\sqrt{119} \mathrm{cm}$
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Question 61 Mark
How many tangents can a circle have?
Answer
A circle can have infinitely many tangents since there are infinitely many points on the circumference of the circle and at each point of it, it has a unique tangent.
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Question 71 Mark
Prove that in two concentric circles, the chord of the larger circle, which touches the smaller circle is bisected at the point of contact.
Answer

In larger circle $C1 , AB$ is the chord and $OP$ is the tangent.
Therefore, $\angle OPB=90^\circ$
Hence, $AP = PB$ ( perpendicular from center of the circle to the chord bisects the chord)
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1 Marks Question - Maths STD 10 Questions - Vidyadip