Download App

Maths 2 Marks Questions

Circles · Maharashtra Board STD 10 (MSBSHSE - English Medium). 21 questions.

Questions

2 Marks Questions

Take a timed test

21 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks
What is the distance between two parallel tangents of a circle of radius 4cm?
Answer

PQ and RS are parallel length.
we know radius always perpendicular to tangent.
So, we say distance between two parallel tangents is equal to diameter.
Then, Diameter = 2 × radius.
Diameter = 2 × 4cm
Diameter = 8cm
Hence, Distance between two parallel tangents are 8cm.
View full question & answer→
Question 22 Marks
In the figure, BOA is a diameter of a circle and the tangent at a point P meets BA produced at T. If $\angle\text{PBO}=30^{\circ}$ then find $\angle\text{PTA}.$
Answer
As, $\angle\text{BPA}=90^{\circ}$
$\angle\text{PAB}=\angle\text{OPA}=60^{\circ}$
Also $\text{OP}\bot\text{OT}$
Therefore,$\angle\text{APT}=30^{\circ}$
and $\angle\text{PTA}=60^{\circ}-30^{\circ}=30^{\circ}$
View full question & answer→
Question 32 Marks
In the figure, PQL and PRM are tangents to the circle with centre O at the points Q and R respectively and S is a point on the circle such that $\angle\text{SQL}=50^{\circ}$ and
$\angle\text{SRM}=60^{\circ}.$ Then, find $\angle\text{QSR}.$
Answer
Here $\angle\text{OSQ}=\angle\text{OQS}=90^{\circ}-50^{\circ}=40^{\circ}$
and $\angle\text{RSO}=\angle\text{SRO}=90^{\circ}-60^{\circ}=30^{\circ}$
Therefore, $\angle\text{QSR}=40^{\circ}+30^{\circ}=70^{\circ}$
View full question & answer→
Question 42 Marks
In the figure, CP and CQ are tangents from an external point C to a circle with centre O. AB is another tangent which touches the circle at R. If CP = 11cm and BR = 4cm, find the length of BC.
Answer
CP and CQ are the tangents to the circle from C.
AB is another tangent to the same circle which touches at R and meets the first two tangents at A and B. O is the centre of the circle.
OC is joined
CP = 11cm, BR = 4cm
CP and CQ are tangents to the circle
CP = CQ = 11cm
Similarly from B, CR and BQ are the tangents
BQ = BR = 4cm
Now BC = CQ – BQ = 11 – 4 = 7cm
View full question & answer→
Question 52 Marks
In the figure, PA and PB are tangents to the circle drawn from an external point P. CD is a third tangent touching the circle at Q. If PB = 10cm and CQ = 2cm, what is the length PC?
Answer
In the figure, PA and PB are the tangents to the circle drawn from P.
CD is the third tangent to the circle drawn at Q.
PB = 10cm, CQ = 2cm
PA and PB are tangents to the circle.
PA = PB = 10 cm
Similarly CQ and CA are tangents to the circle.
CQ = CA = 2cm
PC = PA – CA = 10 – 2 = 8cm.
View full question & answer→
Question 62 Marks
If PT is a tangent at T to a circle whose centre is O and OP $= 17\ cm$, OT $= 8\ cm$, Find the length of the tangent segment PT.
Answer


Given,
OT $= 8\ cm$
OP $= 17\ cm$
Now by PGT in $\triangle\text{OTP}$
$(OP)^2 = (OT)^2 + (PT)^2$
$(17)^2 = (8)^2 + (PT)^2$
$289 - 64 + (PT)^2$
$289 - 64 = (PT^{)2}$
$(PT)2 = 225$
$\text{PT}=\sqrt{225}$
PT $= 15\ cm$.
View full question & answer→
Question 72 Marks
In the figure, $\triangle\text{ABC}$ is circumscribing a circle. Find the length of BC.
Answer
$\triangle\text{ABC}$ is circumscribing a circle which touches it at P, Q and R.
AC = 11cm, AR = 4cm, BR = 3cm
Now we have to find BC.
AR and AQ are tangents to the circle from A.
AQ = AR = 4cm
Then CQ = AC – AQ = 11 – 4 = 7cm
Similarly,
CP and CQ are tangents from C.
CP = CQ = 7cm
and BP and BR are tangents from B.
BP = BR = 3cm
Now BC = BP + CP = 3 + 7 = 10cm.
View full question & answer→
Question 82 Marks
If from any point on the common chord of two intersecting circles, tangents be drawn to the circles, prove that they are equal.
Answer
Given: QR is the common chord of two circles intersecting each other at Q and R. P is a point on RQ when produced From PT and RS are the tangents drawn to tire circles with centres O and C respectively
.
To prove: $PT = PS$ Proof: PT is the tangent and PQR is the secant to the circle with centre O.
$PT^2 = PQ \times PR ….$(i) Similarly PS is the tangent and PQR is the secant to the circle with centre C.
$PS^2 = PQ \times PR ….$(ii) From (i) and (ii) $PT^2 = PS^2 PT = PS$ Hence proved.
View full question & answer→
Question 92 Marks
If the sides of a quadrilateral touch a circle. prove that the sum of a pair of opposite sides is equal to the sum of the other pair.
Answer
we have to prove sum of a pair of opposite sides is equal to sum of other pair.
First, we consider, AB + DC there AB + DC = AF + FB + DH + HC from the property of tangent. we know, AF = AE FB = BG DH = DE HC = CG By putting these value in eq in (i), we get ⇒ AB + DC = AE + BG + DE + CG ⇒ AB + DC = (AE + DE) + (BG + CG) ⇒ AB + DC = AD + BC we have proved that sum of pair of opposite sides is equal to sum of other pair.
View full question & answer→
Question 102 Marks
Two tangents TP and TQ are drawn from an external point T to a circle with centre O . If they are inclined to each other at an angle of 100°, then what is the value of $\angle\text{POQ}$?
Answer
Consider the quadrilateral OPTQ. It is given that $\angle\text{PTQ}=100^{\circ}$ From the property of the tangent we know that the tangent will always be perpendicular to the radius at the point of contact. Therefore we have, $\angle\text{OQT}=90^{\circ}$ $\angle\text{OPT}=90^{\circ}$ We know that the sum of all angles of a quadrilateral will always be equal to 360°. Therefore,$\angle\text{PTQ}+\angle\text{OQT}+\angle\text{OPT}+\angle\text{POQ}=360^{\circ}$
Let us substitute the values of all the known angles. We have, $100^\circ+ 90^\circ+ 90^\circ+\angle\text{POQ}=360^{\circ}$$280 + \angle\text{POQ}=360^{\circ}$
$\angle\text{POQ}=80^{\circ}$
Therefore, the value of angle$\angle\text{POQ}\ \text{is}\ 80^{\circ}.$
View full question & answer→
Question 112 Marks
In the given figure, PA and PB are tangents to the circle with centre O such that $\angle\text{APB}=50^{\circ}$. Write the measure of $\angle\text{OAB}.$
Answer
In the given figure, PA and PB are tangents to the circle from P. PA = PB$\angle\text{APB}=50^{\circ},$ OA is joined.
To find $\angle\text{OAB}$ In$\triangle\text{PAB}$ PA = PB $\therefore\angle\text{PAB}=\angle\text{PBA}$ $\therefore\angle\text{PAB}=\frac{180^{\circ}-\angle\text{APB}}{2}=\frac{180^{\circ}-50^{^{\circ}}}{2}$ $=\frac{130^{\circ}}{2}=65^{\circ}$ But $\angle\text{OAP}=90^{\circ}$ $\therefore\angle\text{OAB}=90^{\circ}-\angle\text{PAB}$$=90^\circ-65^\circ=25^\circ$
View full question & answer→
Question 122 Marks
What the distance between two parallel tangents to a circle of radius 5cm?
Answer

PQ and RS are parallel tangent.
we know radius always perpendicular to tangent.
So, we say distance between two parallel tangent is equal to diameter.
then, Diameter = 2 × radius.
Diameter = 2 × 5cm
Diameter = 10cm
Hence, Distance between two parallel tangents are 10cm.
View full question & answer→
Question 132 Marks
In the figure, PA and PB are tangents to the circle drawn from an external point P. CD is a third tangent touching the circle at Q. If PB = 10cm and CQ = 2cm, what is the length PC ?, if PB = 10cm, what is the perimeter of $\triangle\text{PCD}$?
Answer

By the property of tangent
PA = PB = 10cm tangent from point P
CA = CQ (tangent from point C)
DQ = DB (tangent from point D)
Given, PB = 10cm
Then,
Perimeter of $\triangle\text{PCD}$ = PC + CD + PD
Perimeter of $\triangle\text{PCD}$ = PC + CQ + QD + PD
Perimeter of $\triangle\text{PCD}$ = PC + CA + DB + PD
Perimeter of $\triangle\text{PCD}$ = PA + PB
Perimeter of $\triangle\text{PCD}$ = 2 PB
Perimeter of $\triangle\text{PCD}$ = 2 × 10 = 20cm
Hence, Perimeter of $\triangle\text{PCD}$ is 20cm.
View full question & answer→
Question 142 Marks
How many tangents can a circle have?
Answer
Tengent: A line intersecting circle in one point is called a tangent.
As there are infinite number of point on the circle a circle has many(infinite) tangents.
View full question & answer→
Question 152 Marks
In the given figure, CP and CQ are tangents to a circle with centre O. ARB is another tangent touching the circle at R. If CP = 11cm and BC = 7cm, then find the length of BR.

 
Answer
By the property of tangent
CP = CQ = 11cm (tangent from point c)
AP = AR (tangent from point A)
BQ = BR (tangent from point B)
CB = 7 (given)
Now, we have to find BR,
BQ = CQ - CB
⇒ BR = 11 - 7
⇒ BR = 4cm
Hence, length of BR is 4cm.
View full question & answer→
Question 162 Marks
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.
Answer
We know that,

The at point of contact, the tangent is perpendicular to the radius. Radius is line from center to point on circle. Therefore, perpendicular to tangent will pass
through center of the circle.
View full question & answer→
Question 172 Marks
In the figure, there are two concentric circles with centre O. PRT and PQS are tangents to the inner circle from a point P lying on the outer circle. If PR = 5cm, find the
lengths of PS.
Answer
Given that PR = 5cm.
PR and PQ are the tangents to the inner circle so,
PR = PQ = 5cm (Tangents drawn from an external point to the circle are equal)
Now draw a perpendicular from the centre O to the tangent PS.
PS is the chord of the inner circle. we know that the perpendicular drawn from the centre of the circle to the chord bisects the chord. So, PQ = QS = 5cm.
PS = PQ + QS = 5cm + 5cm = 10cm
View full question & answer→
Question 182 Marks
Find the length of a tangent drawn to a circle with radius $5\ cm$, from a point $13\ cm$ from the centre of the circle.
Answer


Given,
$OA = 5\ cm$
$OB = 13\ cm$
we know, $AB = x\  cm$
radius OT is always perpendicular to tangent TP
By using pythagoras, we find OP
$OB^2 = OA^2 + AB^2$
$\Rightarrow (13)^2= (5)^2 + (x)^2$
$^\Rightarrow 169 = 25 + x^2$
$\Rightarrow x^2 = 169 - 25$
$\Rightarrow\text{x}=\sqrt{144}$
Hence, length of $OA$ is $12\ cm$.
View full question & answer→
Question 192 Marks
Two circles touch externally at a point P. From a point T on the tangent at P, tangents TQ and TR are drawn to the circles with points of contact Q and R respectively.Prove that TQ = TR.
Answer

By using property of tangent from external point,
TQ = TP ...(i)
TR = TP ...(ii)
From eq. (i) and eq. (ii)
⇒ TR = TQ,
Hence proved.
View full question & answer→
Question 202 Marks
A point P is $26\ cm$ away from the centre O of a circle and the length PT of the tangent drawn from P to the circle is $10\ cm$. Find the radius of the circle.
Answer
From a point P outside the circle of centre O and radius OT, PT is the tangent to the circle $OP = 26\ cm, PT = 10\ cm$.

Now in right $\triangle\text{OPT}$ Let r be the radius.
$OP^2 = OT^2 + PT^2$​​​​​​​ (Pythagoras Theorem)
$\Rightarrow (26)^2 = r^2 + (10)$
$ \Rightarrow 676 = r^2 + 100$
$ \Rightarrow 676 – 100 = r^2 $
$\Rightarrow r^2 = 576 = (24)^2 r = 24$
 Hence radius of the circle $= 24\ cm$
View full question & answer→
Question 212 Marks
The length of tangent from a point A at a distance of $5\ cm$ from the centre of the circle is $4\ cm$. What is the radius of the circle?
Answer
$PA$ is a tangent to the circle from $P$ at a distance of $5\ cm$ from the centre $O$.
$PA = 4\ cm$
OA is joined and let $OA = r$.

Now in right $\triangle\text{OAP},$
$OP^2 = OA^2 + PA^2$
$\Rightarrow (5)^2 = r^2 + (4)^2$
$\Rightarrow 25 = r^2 + 16$
$\Rightarrow r^2 = 25 – 16 = 9 = (3)^2$
$r = 3$
Radius of the circle $= 3\ cm$.
View full question & answer→
Generate Paper FreeAll Circles topics