Class 10 Maths - Circles

Draw two tangents from a point $5$ cm away from the centre of a circle of radius $3$ cm.

Given, a circle of radius

$3$ cm whose Centre is

$O$ and a point P

$, 5$ cm away from its centre.

STEPS OF CONSTRUCTION:

$1.$ Draw circle with

$O$ as centre and radius

$= 3 cm.$ Take a point

$P$ such that

$OP= 5 cm.$

$2.$ Draw the bisector of

$OP$ which intersect

$OP$ at

$M.$

$3.$ Taking

$M$ as centre and

$MO$ as radius, draw a dotted circle. Let this circle cuts the given circle

$A$ and

$B.$

$4.$ Join

$PA$ &

$PB$ .

$5.$ Thus ,

$PA$ &

$PB$ are the required tangents. By measurement using scale

$PA= PB = 4 cm.$

1316380_1068946_ans_8b7409dc3d2e4a1383c37f9350ce6d1b.png

Maximum number of tangents that can be drawn from an external point to a circle is ______ .

The tangent of a circle is a special case of the secant when the two end points of the corresponding chord coincide.

From an external point

$P$ , only two tangents (

$PA$ and

$PB$ ) can be drawn.

Also, both the drawn tangents are equal in length.

Hence, maximum

$2$ tangents can be drawn from an external point to a circle.

How many tangents can be drawn on the circle of radius $5\ \text{cm}$ from a point lying outside the circle at distance $9\ \text{cm}$ from the center.

Number of tangents that can be drawn to a circle from a point lying outside the circle is two.

Since, distance between the center and the point (

$9\ \text{cm}$ ) is greater than the radius (

$5\ \text{cm}$ ), the point lies outside the circle. Hence, two tangents can be drawn on the circle.

i) A circle can have _ parallel tangents.
ii) The point common to the tangent and the circle is called ___.

i) A circle can have two parallel tangents.

ii) The point common to the tangent and the circle is called the point of contact.

What is the locus of the centre of a wheel of a bicycle going straight along a level road.

The locus of the centre of a wheel, which is going straight along a level road will be a straight line parallel to the road at a distance equal to the radius of the wheel.
1796986_1832813_ans_c626dd938aa74db7b3ade64d590f75d1.png

1796986_1832813_ans_c626dd938aa74db7b3ade64d590f75d1.png

Find the equation of the tangent to the circle $\displaystyle x^{2}+y^{2}-30x+6y+109=0$ at (4, -1)

Equation of tangent is

$\displaystyle 4x+(-y)-30\left ( \frac{x+4}{2} \right )+6\left ( \frac{y+(-1)}{2} \right )+109=0$
or

$4x - y - 15x - 60 + 37 - 3 + 109 = 0 or -11x + 2y + 46 = 0$
or

$11x - 2y - 46 = 0$
Hence the required equation of the tangent is

$11x - 2y - 46 = 0$

The number of tangents that can be drawn to a circle at a given point on it is

Consider given the two dice,

If the point is on the circle then number of tangent $= 1$

If the point is outside the circle then number of tangent $=2$

If the point is in the circle then number of tangent $=$ none

Hence, this is the answer.

From an external point P, two tangents, PA and PB are drawn to a circle with centre O. At one point E on the circle tangent is drawn which intersects PA and PB at C and D, respectively. If PA = 10 cm, find the perimeter of the triangle PCD in cm.

From the graph, we find some conclusion-

$PA=PB=10$ cm

$CE=CA$

$DE=DB$

Perimeter of

$\Delta PCD$ is,

$=PC+PD+CD$

$\Rightarrow PC+PD+CE+ED$

$\Rightarrow PC+PD+CA+DB$

$\Rightarrow (PC+CA)(PD+DB)$

$\Rightarrow PA+PB$

$\Rightarrow 10+10$

$\Rightarrow 20$ cm.

1458599_427348_ans_7797fef64bd241dcb1f1f40ac46b37a1.png

In the ddjoining figure $O$ is the centre of the circle. Forom point $R\ seg\ RM$ and $seg\ RN$ are tengent segment touching the circle at $M$ and $N$ . If $(OR)=10\ cm$ and radius of the circle $=5\ cm$ , then
What is the measure os $\angle MRN$ ?

We have

$OM=5$ cm ad

$OR=10$ cm

In rightangled

$\triangle{OMR},$

${OR}^{2}={OM}^{2}+{MR}^{2}$

$\Rightarrow MR=\sqrt{{OR}^{2}-{OM}^{2}}=\sqrt{{10}^{2}-{5}^{2}}=\sqrt{100-25}=\sqrt{75}=5\sqrt{3}$ cm

Tangent segments drawn from an external point to a circle are congruent.

$\therefore MR=NR=5\sqrt{3}$ cm

In right angled

$\triangle{OMR}$

$\tan{\angle{MRO}}=\dfrac{OM}{MR}$

$=\dfrac{5}{5\sqrt{3}}=\dfrac{1}{\sqrt{3}}$

$\Rightarrow \tan{\angle{MRO}}=\tan{{30}^{\circ}}$

Thus, the measure of

$\angle{MRO}$ is

${30}^{\circ}$

In right angled

$\triangle{ONR}$

$\tan{\angle{NRO}}=\dfrac{ON}{NR}$

$=\dfrac{5}{5\sqrt{3}}=\dfrac{1}{\sqrt{3}}$

$\Rightarrow \tan{\angle{NRO}}=\tan{{30}^{\circ}}$

Thus, the measure of

$\angle{NRO}$ is

${30}^{\circ}$

$\therefore \angle{MRN}=\angle{MRO}+\angle{NRO}={30}^{\circ}+{30}^{\circ}={60}^{\circ}$

Thus, the measure of

$\angle{MRN}$ is

${60}^{\circ}$

If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that $\angle DBC = {120^0}$ prove that BC+BD=BO, i.e, BO=2BC.

Sol. Given : A circle with centre O.

Tangents BC and BD are drawn from an external point B such that

$\angle DBC = 120^{\circ}$

To prove :

$BC + BD = BO,$ i.e.,

$BO = 2BC$

Construction : join

$OB, OC$ and

$OD$ .

Proof : In

$\Delta OBC$ and

$\Delta OBD$ , we have

$OB = OB$ [common]

$OC = OD$ [Raddi of same circle]

$BC = BD$ [Tangents from an external point are equal in length ] __(i)

$\therefore \Delta OBC \cong \Delta OBD$ [By SSS criterion of congruence]

$\Rightarrow \angle OBC = \angle OBD (CPCT)$

$\therefore \angle OBC = \dfrac{1}{2} \angle DBC = \dfrac{1}{2} \times 120^{\circ}$ [

$\because \angle CBD = 120^{\circ}$ given]

$\Rightarrow \angle OBC = 60^{\circ}$

OC and BC are radius and tangent respectively at contact point C.

So,

$\angle OCB = 90^{\circ}$

Now, in right angle

$\Delta OBC = 60^{\circ}$

$\therefore cos 60^{\circ} = \dfrac{BC}{BO}$

$\Rightarrow \dfrac{1}{2} = \dfrac{BC}{BO}$

$\Rightarrow OB = 2 BC$

Hence, proved (ii) part,

$\Rightarrow OB = BC + BC$

$\Rightarrow OB = BC + BD$ [

$\because BC =BD$ from (i)]

Hence, proved.

1791821_1458800_ans_255966551fee44788fd0f37fae1b9ede.png

Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60 .

${\textbf{Step - 1: Finding various angles}}{\text{.}}$

${\text{As we have to make to angle between the tangents equal to 60}}^\circ$

${\text{and the tangents subtend an angle of 90}}^\circ {\text{ at its point of contact}}{\text{.}}$

${\text{We can write, that the sum of all the angles of a quadrilateral is equal to 360}}^\circ {\text{,}}$

${\text{So for the quadrilateral we have the three angles equal to 60}}^\circ ,90^\circ ,{\text{and 90}}^\circ$

${\text{Therefore,the angle subtended by the pair of tangents at the centre will be }}$

$360^\circ - 90^\circ - 90^\circ - 60^\circ = 120^\circ$

${\textbf{Step - 2: Drawing the figure}}{\text{.}}$

${\textbf{Hence, the pair of the tangents can be drawn as above}}{\text{.}}$

Two tangents PA and PB are drawn to a circle with centre O, such that $\angle APB=120^\circ$ Prove that $OP=AP+BP=2AP$

Given: O is the centre of the circle. PA and PB are tangents drawn to a circle and ∠APB = 120°.

To prove: OP = 2AP

Proof:

In ΔOAP and ΔOBP,

OP = OP (Common)

∠OAP = ∠OBP (90°) (Radius is perpendicular to the tangent at the point of contact)

OA = OB (Radius of the circle)

∴ ΔOAP is congruent to ΔOBP (RHS criterion)

∠OPA = ∠OPB = 120°/2 = 60° (CPCT)

In ΔOAP,

cos∠OPA = cos 60° = AP/OP

Therefore, 1/2 =AP/OP

Thus, OP = 2AP

Hence, proved.

A circle touches the side $BC$ of $\triangle\ ABC$ at $P$ and touches $AB$ and $AC$ produced at $Q$ and $R$ respectively. Prove that $AQ=\dfrac{1}{2}$ (perimeter of $\triangle\ ABC$ ).

Given:A circle touching the side

$BC$ of

$\triangle{ABC}$ at

$P$ and

$AB$ ,

$AC$ produced at

$Q$ and

$R$ respectively.

Proof: Lengths of tangents drawn from an external point to a circle are equal.

$\Rightarrow AQ = AR, BQ = BP, CP = CR$ .

Perimeter of

$\triangle{ABC}=AB + BC + CA$

$= AB +\left(BP + PC\right)+\left(AR – CR\right)$

$=\left(AB + BQ\right)+\left(PC\right)+\left(AQ – PC\right)$ since

$\left[AQ = AR, BQ = BP, CP = CR\right]$

$= AQ + AQ$

$= 2AQ$

$\Rightarrow AQ =\dfrac{1}{2}$ (Perimeter of

$\triangle{ABC}$ )

$\therefore AQ$ is the half of the perimeter of

$\triangle{ABC}$

Write the name of the common point of the tangent to a circle and the circle.

The common point of the tangent to a circle and the circle is called the point of contact.

Which term will you use for a line that intersects a circle at two distinct points?

The term used for a line that intersects a circle at two distinct points is secant. When a line touches a circle at only one point it is known as tangent.

How many tangents can a circle have?

${\textbf{Step1: Find the number of tangents to the circle from the given points. }}$

${\text{ Circle is the locus of the points equidistant from given points, which is the center of the circle }}$

${\text{and tangent is the line that intersects a circle at one point only.}}$

${\text{ As circle has infinite points so there will be infinite tangents can be drawn on these points}}$

${\text{ which touches at only one point.}}$

${\textbf{Therefore, a circle can have an infinite number of tangents .}}$

$DA$ and $DB$ are the triangle drwan from D, to the circle with centre $O$ , at A and B respectively $\angle ADO = 35^\circ$ find $\angle AOB$

We are given,

Circle of centre

$O$ is having radius

$OA$ and

$OB$ (r)

Now, In

$\Delta OAD$ &

$\Delta OBD$ is equal as two sides are equal and one angle is equal

$(OA = OB = r)$ and

$OD$ also

$m\angle OAD = m \angle OBD = 90^o$

Now, given that

$m\angle OBD = 90^o$

Now, given that

$m \angle ADO = 35^o$

then

$m\angle AOD = ?$

$\Delta AOD, m \angle AOD + m \angle ADO + m \angle OAD = 180^o$

$m\angle AOD = 180 - m \angle OAD - m \angle ADO$

$m \angle AOD = 180 - 90 - 35 = 55^o$

So,

$m\angle AOB = m \angle AOD + m \angle DOB$

$= 55^o + 55^o$

$m \angle AOB = 110^o$

1493219_1529218_ans_eb99b5c0f81b4e1fa8d5799c21dbd6a0.png

Pair of tangents are drawn from point (1,3) to circle ${ x }^{ 2 }+{ y }^{ 2 }-3x-x+3y-1=0$ Find line joining their point of contact

Equation of the circle is given as,

$x^2+y^2-3x-x+3y-1=0$ ……………..

$(1)$

$\Rightarrow x^2+y^2-4x+3y-1=0$

$\Rightarrow x^2+y^2+2(-2)x+2.\left(\dfrac{3}{2}\right)y+(-1)=0$

$\therefore g=-2$ ,

$f=\dfrac{3}{2}$ and

$c=-1$

Equation of the chord of contact of circle

$(1)$ is given as

$x_1x+y_1y+g(x_1+x)+f(y_1+y)+c=0$

$\Rightarrow x_1x+y_1y+(-2)(x_1+x)+\dfrac{3}{2}(y_1+y)-1=0$

Now, equation of the chord of contact of circle

$(1)$ wrt to

$(1, 3)$

$1x+3y+(-2)(1+x)+\dfrac{3}{2}(3+y)-1=0$

$2x+6y-4(1+x)+3(3+y)-2=0$

__________________________________

$2$

$\Rightarrow 2x+6y-4-4x+9+3y-2=0$

$\Rightarrow -2x+9y+3=0$

$\Rightarrow 2x-9y-3=0$

$\Rightarrow 2x-9y=3$ .

1215675_1300969_ans_2cb011034e1b48248997ae8fa3fcdb78.JPG

How many tangents can be drawn at a point on a circle ?

Only one tangent can be drawn from a point on the circle
Every other line intersect the circle at two points which won't be a tangent
1795328_1815673_ans_c6f48f0fac564444a03d703a9f5a5211.png

1795328_1815673_ans_c6f48f0fac564444a03d703a9f5a5211.png

Draw two concentric circles of radii $3 \,cm$ and $5 \,cm.$ Taking a point on outer circle construct the pair of tangents to the other. Measure the length of a tangent and verify it by actual calculation.

Steps of construction:

(i) Draw two concentric circles

$C_1, C_2$ of radii

$3 \,cm$ and

$5 \,cm$ respectively taking 'O' as centre.

(ii) Draw perpendicular bisector AB of OT. T is any point on

$C_2.$

(iii) Draw circle

$C_3$ taking radius

$TM = OM$ and

$M$ as centre.

(iv) Circle

$C_3$ intersect the circle

$C_1$ at P and Q. Join TP and TQ. These are the required tangents.

$TP = TQ = 4.1 \,cm$ by measuring.

Mathematically length of tangent: Join OP. OP and TP are radius and tangent respectively at contact point P. So,

$\angle TPO = 90^\circ.$

By Pythagoras theorem in

$\Delta TPO,$

$PT^2 = OT^2 - OP^2 = 5^2 - 3^2 = 25 - 9 = 16$

$\Rightarrow PT = 4 \,cm$

Difference in measurement and by mathematical calculation

$PT = 4.1 \,cm - 4 \,cm = 0.1 \,cm.$

In figure, the sides AB, BC and CA of triangle ABC touch a circle with centre 0 and radius r at P. Q and R respectively.
Prove That :
(i) AB+CQ=AC+BQ
(ii) Area $(\Delta ABQ=\frac{1}{2}$ (Perimeter of $(\Delta ABC)\times r$

2173000_1582989_ans_71369e9dcad64cb3a2802c108f504b34.jpeg

In the given figure, tangents $PQ$ and $PR$ are drawn from an external point p to a circle with center $O$ , such that $\angle RPQ=30^{\circ}$ . A chord $RS$ is drawn parallel to the tangent $PQ$ . Find $\angle RQS.$

1570510_971375_ans_99a8ec3d198147ca917a817416ee48b4.jpg

In the given figure, O is the centre of a circle of radius 5 cm, T is a point such that OT= 13 cm and OT intersects the circle at E. If AB is the tangent to the circle at E, find the length of AB.

$OP = OQ = 5 cm$

$OT = 13 cm$

$OP$ and

$PT$ are radius and tangent respectively at contact point P.

$\therefore \angle OPT = 90^{\circ}$

So, by pythagoras theorem, in right angled

$\Delta OPT$ ,

$PT^2 = OT^2 - OP^2 = 13^2 - 5^2 = 169 - 25 = 144$

$\Rightarrow PT = 12 cm$

$AP$ and

$AE$ are two tangents from an external point A to a circle.

$\therefore AP = AE$

$AEB$ is tangent and

$OE$ is radius at contact point E.

So,

$AB \bot OT$ ___(i)

So, by Pythagoras theorem, in right angled.

$\Delta AET$ .

$AE^2 = AT^2 - ET^2$

$\Rightarrow AE^2 = (PT - PA)^2 - [TO - OE]^2$

$=(12 - AE)^2 - (13 - 5)^2$

$\Rightarrow AE^2 = (12)^2 + (AE)^2 - 2 (12) (AE) - (8)^2$

$\Rightarrow AE^2 - AE^2 + 24 AE = 144 - 64$

$\Rightarrow 24 AE = 80$

$\Rightarrow AE = \dfrac{80}{24} cm$

$\Rightarrow AE = \dfrac{10}{3} cm$

REF. IMAGE

$\Delta TPO$ and

$\Delta TQO$ ,

$OT = OT$ [common]

$PT = QT$ [Tangents from T]

$OP = OQ$ [Radii of same circle]

$\therefore \Delta TPO \cong \Delta TQO$ [By SSS criterion of congruence]

$\Rightarrow \angle 1 = \angle 2$ ___(ii) [CPCT]

$\Delta ETA$ and

$\Delta ETB$ ,

$ET = ET$ [Common]

$\angle TEA = \angle TEB = 90^{\circ}$ [From (i)]

$\angle 1 = \angle 2$ [CPCT] [From (ii)]

$\therefore \Delta ETA \cong \Delta ETB$ [By ASA criterion of congruence]

$\Rightarrow AE = BE$ [CPCT]

$\Rightarrow AB = 2 AE = 2 \times \dfrac{10}{3}$

$\Rightarrow AB = \dfrac{20}{3} cm$

Hence, the required length is

$\dfrac{20}{3} cm$ .

1804487_1516190_ans_a7fea1e0002d4770b10122e7a1979cb7.png

Circles - Class 10 Maths - Extra Questions

Draw two tangents from a point $5$ cm away from the centre of a circle of radius $3$ cm.

Maximum number of tangents that can be drawn from an external point to a circle is ______ .

How many tangents can be drawn on the circle of radius $5\ \text{cm}$ from a point lying outside the circle at distance $9\ \text{cm}$ from the center.

i) A circle can have _ parallel tangents.
ii) The point common to the tangent and the circle is called ___.

What is the locus of the centre of a wheel of a bicycle going straight along a level road.

Find the equation of the tangent to the circle $\displaystyle x^{2}+y^{2}-30x+6y+109=0$ at (4, -1)

The number of tangents that can be drawn to a circle at a given point on it is

From an external point P, two tangents, PA and PB are drawn to a circle with centre O. At one point E on the circle tangent is drawn which intersects PA and PB at C and D, respectively. If PA = 10 cm, find the perimeter of the triangle PCD in cm.

In the ddjoining figure $O$ is the centre of the circle. Forom point $R\ seg\ RM$ and $seg\ RN$ are tengent segment touching the circle at $M$ and $N$ . If $(OR)=10\ cm$ and radius of the circle $=5\ cm$ , then
What is the measure os $\angle MRN$ ?

If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that $\angle DBC = {120^0}$ prove that BC+BD=BO, i.e, BO=2BC.

Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60 .

Two tangents PA and PB are drawn to a circle with centre O, such that $\angle APB=120^\circ$ Prove that $OP=AP+BP=2AP$

A circle touches the side $BC$ of $\triangle\ ABC$ at $P$ and touches $AB$ and $AC$ produced at $Q$ and $R$ respectively. Prove that $AQ=\dfrac{1}{2}$ (perimeter of $\triangle\ ABC$ ).

Write the name of the common point of the tangent to a circle and the circle.

Which term will you use for a line that intersects a circle at two distinct points?

How many tangents can a circle have?

$DA$ and $DB$ are the triangle drwan from D, to the circle with centre $O$ , at A and B respectively $\angle ADO = 35^\circ$ find $\angle AOB$

Pair of tangents are drawn from point (1,3) to circle ${ x }^{ 2 }+{ y }^{ 2 }-3x-x+3y-1=0$ Find line joining their point of contact

How many tangents can be drawn at a point on a circle ?

Draw two concentric circles of radii $3 \,cm$ and $5 \,cm.$ Taking a point on outer circle construct the pair of tangents to the other. Measure the length of a tangent and verify it by actual calculation.

In figure, the sides AB, BC and CA of triangle ABC touch a circle with centre 0 and radius r at P. Q and R respectively.
Prove That :
(i) AB+CQ=AC+BQ
(ii) Area $(\Delta ABQ=\frac{1}{2}$ (Perimeter of $(\Delta ABC)\times r$

In the given figure, tangents $PQ$ and $PR$ are drawn from an external point p to a circle with center $O$ , such that $\angle RPQ=30^{\circ}$ . A chord $RS$ is drawn parallel to the tangent $PQ$ . Find $\angle RQS.$

In the given figure, O is the centre of a circle of radius 5 cm, T is a point such that OT= 13 cm and OT intersects the circle at E. If AB is the tangent to the circle at E, find the length of AB.

Class 10 Maths Extra Questions

Circles - Class 10 Maths - Extra Questions

Draw two tangents from a point 55 cm away from the centre of a circle of radius 33 cm.

Maximum number of tangents that can be drawn from an external point to a circle is ______ .

How many tangents can be drawn on the circle of radius 5 cm5\ \text{cm} from a point lying outside the circle at distance 9 cm9\ \text{cm} from the center.

i) A circle can have ___ parallel tangents.ii) The point common to the tangent and the circle is called _____.

What is the locus of the centre of a wheel of a bicycle going straight along a level road.

Find the equation of the tangent to the circle x2+y2−30x+6y+109=0\displaystyle x^{2}+y^{2}-30x+6y+109=0 at (4, -1)

The number of tangents that can be drawn to a circle at a given point on it is

From an external point P, two tangents, PA and PB are drawn to a circle with centre O. At one point E on the circle tangent is drawn which intersects PA and PB at C and D, respectively. If PA = 10 cm, find the perimeter of the triangle PCD in cm.

In the ddjoining figure OO is the centre of the circle. Forom point R seg RMR\ seg\ RM and seg RNseg\ RN are tengent segment touching the circle at MM and NN. If (OR)=10 cm(OR)=10\ cm and radius of the circle =5 cm=5\ cm, thenWhat is the measure os ∠MRN\angle MRN?

If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that ∠DBC=1200\angle DBC = {120^0} prove that BC+BD=BO, i.e, BO=2BC.

Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60 .

Two tangents PA and PB are drawn to a circle with centre O, such that ∠APB=120∘\angle APB=120^\circ Prove that OP=AP+BP=2APOP=AP+BP=2AP

A circle touches the side BCBC of △ ABC\triangle\ ABC at PP and touches ABAB and ACAC produced at QQ and RR respectively. Prove that AQ=12AQ=\dfrac{1}{2} (perimeter of △ ABC\triangle\ ABC).

Write the name of the common point of the tangent to a circle and the circle.

Which term will you use for a line that intersects a circle at two distinct points?

How many tangents can a circle have?

DADA and DBDB are the triangle drwan from D, to the circle with centre OO, at A and B respectively ∠ADO=35∘\angle ADO = 35^\circ find ∠AOB\angle AOB

Pair of tangents are drawn from point (1,3) to circle x2+y2−3x−x+3y−1=0{ x }^{ 2 }+{ y }^{ 2 }-3x-x+3y-1=0 Find line joining their point of contact

How many tangents can be drawn at a point on a circle ?

Draw two concentric circles of radii 3cm3 \,cm and 5cm.5 \,cm. Taking a point on outer circle construct the pair of tangents to the other. Measure the length of a tangent and verify it by actual calculation.

In figure, the sides AB, BC and CA of triangle ABC touch a circle with centre 0 and radius r at P. Q and R respectively.Prove That : (i) AB+CQ=AC+BQ(ii) Area (ΔABQ=12(\Delta ABQ=\frac{1}{2} (Perimeter of (ΔABC)×r(\Delta ABC)\times r

In the given figure, tangents PQPQ and PRPR are drawn from an external point p to a circle with center OO, such that ∠RPQ=30∘\angle RPQ=30^{\circ}. A chord RSRS is drawn parallel to the tangent PQPQ. Find ∠RQS.\angle RQS.

In the given figure, O is the centre of a circle of radius 5 cm, T is a point such that OT= 13 cm and OT intersects the circle at E. If AB is the tangent to the circle at E, find the length of AB.

Class 10 Maths Extra Questions

Support mcqexams.com by disabling your adblocker.

Draw two tangents from a point $5$ cm away from the centre of a circle of radius $3$ cm.

How many tangents can be drawn on the circle of radius $5\ \text{cm}$ from a point lying outside the circle at distance $9\ \text{cm}$ from the center.

i) A circle can have _ parallel tangents.
ii) The point common to the tangent and the circle is called ___.

Find the equation of the tangent to the circle $\displaystyle x^{2}+y^{2}-30x+6y+109=0$ at (4, -1)

In the ddjoining figure $O$ is the centre of the circle. Forom point $R\ seg\ RM$ and $seg\ RN$ are tengent segment touching the circle at $M$ and $N$ . If $(OR)=10\ cm$ and radius of the circle $=5\ cm$ , then
What is the measure os $\angle MRN$ ?

If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that $\angle DBC = {120^0}$ prove that BC+BD=BO, i.e, BO=2BC.

Two tangents PA and PB are drawn to a circle with centre O, such that $\angle APB=120^\circ$ Prove that $OP=AP+BP=2AP$

A circle touches the side $BC$ of $\triangle\ ABC$ at $P$ and touches $AB$ and $AC$ produced at $Q$ and $R$ respectively. Prove that $AQ=\dfrac{1}{2}$ (perimeter of $\triangle\ ABC$ ).

$DA$ and $DB$ are the triangle drwan from D, to the circle with centre $O$ , at A and B respectively $\angle ADO = 35^\circ$ find $\angle AOB$

Pair of tangents are drawn from point (1,3) to circle ${ x }^{ 2 }+{ y }^{ 2 }-3x-x+3y-1=0$ Find line joining their point of contact

Draw two concentric circles of radii $3 \,cm$ and $5 \,cm.$ Taking a point on outer circle construct the pair of tangents to the other. Measure the length of a tangent and verify it by actual calculation.

In figure, the sides AB, BC and CA of triangle ABC touch a circle with centre 0 and radius r at P. Q and R respectively.
Prove That :
(i) AB+CQ=AC+BQ
(ii) Area $(\Delta ABQ=\frac{1}{2}$ (Perimeter of $(\Delta ABC)\times r$

In the given figure, tangents $PQ$ and $PR$ are drawn from an external point p to a circle with center $O$ , such that $\angle RPQ=30^{\circ}$ . A chord $RS$ is drawn parallel to the tangent $PQ$ . Find $\angle RQS.$