435 + Interview Questions in CIRCLE IN GENERAL AWARENESS Page 8 InterviewSolution

351.	In the figure, m ( arc PAQ ) =m (arc RBS ) `=130^(@)` chord PQ has length 10cm then what is the length of chord RS ? Why ?
Answer» RS= 10cm , congruent arcs have their correspondingchords congruent.

Discussion

352.	In the adjoining figure , `O` is the centre of the circle. If `angleACB=25^@`, then find `angleAOB`.
Answer» Correct Answer - `50^@`

Discussion

353.	In the adjoining figure, `O` is the centre of the circle. If `angle BOP=130^@`, then find the value of x.
Answer» Correct Answer - `25^@`

Discussion

354.	`O` is the centre of a circle of diameter AB. If chord `AC=` chord BC, then find the value of `angleABC`.
Answer» Correct Answer - `45^@`

Discussion

355.	In the adjoining figure, if `angleACB=40^@`, `angle DPB=120^@`, then find `angleCBD`.
Answer» Correct Answer - `20^@`

Discussion

356.	In the adjoining figure, AOB is the diameter and `O` is the centre of the circle. If `angle BDC=60^@`, then find the value of `angleABC`.
Answer» Correct Answer - `30^@`

Discussion

357.	Show that if two chords of a circle bisect one another they must bediameters.
Answer» Given-> 2 chords AB and CD O is midpoint of AB and CD To prove->AB and CD are diameter Proof->`/_AOC and /_BOD` `/_AOC=/_BOD` `OA=OB` `OC=OD` `/_AOC cong /_BOD(SAS)` `AC=BD-(1)` `In /_AOD and /_BOC` `/_AOD=/_BOC` `OA=OB` `OC=OD` `/_AOD cong /_BOC(SAS)` `AD=BC-(2)` adding equation 1 and 2 `AC+AD=BC+BD` `CAD=CBD` CD is diameter of circle AB is diameter of circle.

Discussion

358.	In the adjoining figure, `O` is the centre of the circle. If `angle AOB= 150^@` and `angleBOC=100^@`, then find the value of `angleABC`.
Answer» Correct Answer - `55^@`

Discussion

359.	In the adjoining figure, if ABCD is a cyclic quadrilateral, find the value of x.
Answer» Correct Answer - `100^@`

Discussion

360.	In the adjoining figure, `O` is the centre of the circle and AC is its diameter. If `angleBAC=30^@`, then find `angleBOC`.
Answer» Correct Answer - `60^@`

Discussion

361.	In fig. ∠ACB = 40°, find ∠OAB
Answer» ∴ ∠ACB = 40° We know that angle subtended by arc of a circle at center is double the angle at remaining part. ∴ ∠AOB = 2∠ACB = 2 × 40° = 80° ⇒ ∠AOB = 80° ∵ OA = OB = radius of circle ∴ In ∆AOB ∠OAB + ∠OBA + ∠AOB = 180° ∵ Angles opposite to equal sides are equal. ∴ ∠OAB = ∠OBA ⇒ ∠OAB + ∠OAB + 80° = 180° ⇒ 2∠OAB = 180° – 80° ⇒ ∠OAB = \(\frac { { 100 }^{ \circ } }{ 2 } \) ⇒ ∠OAB = 50°

361.

In fig. ∠ACB = 40°, find ∠OAB

Answer»

∴ ∠ACB = 40°

We know that angle subtended by arc of a circle at center is double the angle at remaining part.

∴ ∠AOB = 2∠ACB

= 2 × 40°

= 80°

⇒ ∠AOB = 80°

∵ OA = OB = radius of circle

∴ In ∆AOB

∠OAB + ∠OBA + ∠AOB = 180°

∵ Angles opposite to equal sides are equal.

∴ ∠OAB = ∠OBA

⇒ ∠OAB + ∠OAB + 80° = 180°

⇒ 2∠OAB = 180° – 80°

⇒ ∠OAB = \(\frac { { 100 }^{ \circ } }{ 2 } \)

⇒ ∠OAB = 50°

Discussion

362.	In figure, if ∠OAB = 40°, then ∠ACB equals(a) 50°(b) 40°(c) 60°(d) 70°
Answer» Answer is (a) 50° ∵ OA = OB (radius of circle) ∴ ∠OAB = ∠OBA (angles opposite to equal sites) ⇒ ∠OBA = 40° In right angled ∆OAB ∠OAB + ∠OBA + ∠AOB = 180° ⇒ 40° + 40° + ∠AOB = 180° ⇒ ∠AOB = 180° – 80° ⇒ ∠AOB = 100° We know that ∠ACB = \(\frac { 1 }{ 2 }\)∠AOB = \(\frac { 1 }{ 2 }\) x 100° = 50°

362.

In figure, if ∠OAB = 40°, then ∠ACB equals(a) 50°(b) 40°(c) 60°(d) 70°

Answer»

Answer is (a) 50°

∵ OA = OB (radius of circle)

∴ ∠OAB = ∠OBA (angles opposite to equal sites)

⇒ ∠OBA = 40°

In right angled ∆OAB

∠OAB + ∠OBA + ∠AOB = 180°

⇒ 40° + 40° + ∠AOB = 180°

⇒ ∠AOB = 180° – 80°

⇒ ∠AOB = 100°

We know that

∠ACB = \(\frac { 1 }{ 2 }\)∠AOB

= \(\frac { 1 }{ 2 }\) x 100° = 50°

Discussion

363.	In the adjoining figure, ABCD is a cyclic quadrilateral whose side AB is the diameter of the circle . If `angle ADC=140^(@)` , then find the value of `angleBAC`.
Answer» Since, ABCD is a cyclic quadrilateral . `thereforeangle ADC+angle ABC=180^@` `rArr140^@+angle ABC=180^@` `rArrangle ABC=180^@-140^@` `rArrangle ABC =40^@` AB is the diameter of circle. `thereforeangleACB=90^@` (angle in a semi- circle) Now, in `Delta ABC` ` angle BAC=angle ABC+anlge ACB=180^@` `rArrangle BAC +40^@+90^@=180^@` `rArrangle BAC =180^@-130^@` `rArrangle BAC -50^@`

Discussion

364.	In the adjoining figure, `O` is the centre of the circle. If `angleBOD=50^@`, then find `angleBCD`. (ii) In the adjoining figure,`O` is the centre of the circle. If `angleABC=20^@`, then find `angleAOC`.
Answer» Correct Answer - (i) `25^@` (ii) `40^@`

Discussion

365.	Number of circles passes through three collinear points(a) One(b) Two(c) Zero(d) Infinite
Answer» Answer is (a) One

Discussion

366.	In the figure, two circles intersect each other at points A and B. AP and AQ are the diameters of these circels. Prove that PBQ is a straight line.
Answer» Join AB. In first circle, AP is the diameter of the circle. `therefore angle ABP=90^@` (angle in a semi-circle) In second circle, AQ is the diameter of the circle. `therefore angle ABQ=90^@` Now, `angleABP+angle ABQ=90^@+90^@` `rArrangle PBQ=180^@` `rArrPBQ` is a stright line.

Discussion

367.	In the adjoining figure, ABCD is a cyclic quadrilateral. If `angleDBC=60^(@)` and `angleBAC=40^(@)` , then find the value of `angleBCD` .
Answer» Since, ABCD is a cyclic quadrilateral . `thereforeangle DAC =angle DBC=60^@` and ` angle DAB=angle DAC+angle CAB` `=60^@+40^@` `=100^@` Now, `angle DAB+angle BCD=180^@` `rArr100^@+angle BCD =180^@` `angle BCD =180^@-100^@` `[email protected]` .

Discussion

368.	In the adjoining , `O` is the centre of the circle. `ACB` is a segment. If `angle OAB=30^(@)`, then find the value of `angle ACB`.
Answer» In `Delta OAB`, `AO=BO` (radii of circle) `rArrangle ABO=angleBAO` `rArrangle ABO=30^@` Now, `angle AOB=180^@-(angleABO+angle BAO)` `=180^@-(30^@+30^@)` `120^@` Arc AB subtends angles at centre `=angle AOB` remaining circle `=angle ACB` `therefore angle ACB=(1)/(2)angle AOB` `=(1)/(2)xx120^@` `=60^@`

Discussion

369.	Points A,B,C are on a circle, such that m( arc AB ) = m ( arc BC ) `=120^(@)` . No point, except point B, is common to the arcs. What type is the `Delta ABC `?A. Equilateral triangleB. Scalene triangleC. Right angled triangleD. Isosceles triangle
Answer» Correct Answer - A

Discussion

370.	In the adjoining figure, AC is the diameter of the circle. If `angleBDC=115^(@)`, then find the value of `angle ACB`.
Answer» square `ABDC` is a cyclic quadrilateral. `therefore angle BDC+BAC =180^@` `rArr115^@+angle BAC =180^@` `rArrangle ACB=180^@-115^@` ` 65^@` In `Delta ABC, AC` is the diameter of the circle. `thereforeangle ABC =90^@` Now, `angle ACB=180^@-(angle ABC +angle BAC) ` `=180^@=(90^@+65^@)` `rArrangle ACB =25^@` .

Discussion

371.	In a cyclic `square ` ABCD, twice the measure of `/_A` is thrice the measure of `/_C ` . Find the measure of `/_C `.A. `36^(@)`B. `72^(@)`C. `90^(@)`D. `108^(@)`
Answer» Correct Answer - B

Discussion

372.	In the adjoining figure, `Delta ABC` is an isosceles triangle. Find the value of `angleBDC ` and `angleBEC`.
Answer» Correct Answer - `angleBCD=60^@,angleBEC=120^@`

Discussion

373.	An angle of a cyclic trapezium is twice the other angle. Find the value of the smaller angle.
Answer» Correct Answer - `60^@`

Discussion

374.	If O is the centre of a circle with radius r and AB is a chord of the circle at a distance `r/2` from O, then `/_BAO=`
Answer» Correct Answer - `30^@`

Discussion

375.	Find the radius of the smalles circle which touches the straight line `3x-y=6`at `(-,-3)`and also touches the line `y=x`. Compute up to one place of decimal only.
Answer» Correct Answer - 1.5 unit The equation of the circle touching the line` 3x-y-6=0` at (1,-3) is given by `(x-1)^(2)+(y+3)^(2)+lambda(3x-y-6)=0` i.e., `x^(2)+y^(2)+(3lambda-2)x+(6-lambda)y+10-6 lambda=0` Equation (1) touches `y=x`. Therefore, `2x^(2)+(2lambda+4)x+10-6 lambda =0` i.e., `x^(2)+(lambda+2)x+5-3 lambda=0` has equal roots. Therefore, `(lambda+2)^(2)-4(5-3 lambda)=lambda^(2)+16lambda-16=0` or `lambda=(1)/(2)[-16+- sqrt(256+64)]= - 8+- sqrt(80)` ltbr. Now, `("Radius")^(2)=R^(2)=(1)/(4)[(3lambda-2)^(2)+(6-lambda)^(2)-(10-6lambda)4]` `=(1)/(4)[10lambda^(2)]` or `R=(sqrt(10)lambda)/(2)=(sqrt(10))/(2)(-8+- 4 sqrt(5))=\|-4 sqrt(10)+-10sqrt(2)\|` or Radius of smaller circle `=\|-4 sqrt(10)+10sqrt(2)\|=1.5` approx.

Discussion

376.	`square` PQRS is cyclic .T is a point of ray SR such that S-R-T and `/_QRT 100^(@)` then what is the measure of `/_ SPQ` ? Why ?
Answer» `/_SPQ = 100^(@)`, corollary of cyclic quadrilateral theorem.

Discussion

377.	In adjoining figure, PQ = QR . `angleP = 60^(@)` `:.` m (arc PR)= ..... A. `120^(@)`B. `60^(@)`C. `90^(@)`D. `240^(@)`
Answer» Correct Answer - A

Discussion

378.	In the figure, O is the centre of circle `/_ QPR = 70^(@)` and m (arc PYR ) `= 160^(@)`, then find the value of each of the following `:` (a) m ( arc QXR ) (b) `/_QOR` (c ) `/_ PQR `
Answer» Correct Answer - (a) m ( arc QXR ) 140 (b) `/_ QOR = 140^(@)` (c ) `/_ PQR = 80^(@)` `/_QPR = (1)/(2) ` m (arc QXR ) …(Inscribed angle theorem ) `:. 70 - (1)?92) ` m (arc QXR ) `:. ` m (arc QXR ) `= 70 xx 2 ` `:. ` m (arc QXR `= 140^(@) ` (b) `/_QOR =` m ( arc QXR ) ….(Definition of measure of minor arc ) `:. /_ QOR = 140^(@)` (c ) ` /_PQR = (1)/(2) ` m ( arc PYR ) ....(Inscribed angle theorem ) `:. /_ PQR = (1)/(2) xx 160^(@)` ` :. /_ PQR = 80^(@)`

Discussion

379.	In the figure , `Delta QRS ` is an equilateral triangle. Prove that (1) arc RS `~= arc QS ~= ` arc QR (2) m (arc QRS ) `= 240^(@)`. Proof `:`
Answer» `Delta QRS ` is an equilateral triangle ….(Given ) `:.` seg RS `~= ` seg QS `~=` seg QR …(Sides of equilateral triangle ) Arcs of the same circle are equal, if the related chords are congruent. `:. ` arc RS `~=` arc QS `~=` arc QR ....(1) Let m(arc RS ) = m (arc QS ) = m (arc QR ) = x m ( arc RS ) + m (arc QS ) + m ( arc QR ) ` = 360^(@)` ....( Measure of the circle is `360^(@)` ) `:. x + x + x= 360^(@)` ....[From (1) ] `:. 3x = 360^(@)` `:. x = ( 360 )/( 3)` `:. x = 120^(@)` `:. ` m ( arc RS ) = m (arc QS ) = m (arc QR )` = 120^(@)` m(arc QRS ) = m (arc ( QR 0 + m (arc RS ) ....(Arc addition postulate ) `:.` m ( arc QRS ) `= 120 ^(@) + 120^(@)` `:.` m ( arc QRS ) = `240^(@)`

Discussion

380.	In the figure, P is the cnetre of the circle. If `/_ QPR 50^(@), /_ RPS = 60^(@) , /_ SPT = 100^(@)` then find (1) m(arc QRS) (ii) m ( arc QST ) (iii) m (arc RTS )
Answer» m (arc QRS) = `110^(@)`, m ( arc QST ) `= 210^(@)`, m (arc RTS ) `= 300^(@)`

Discussion

381.	The centre of circleinscribed in a square formed by lines `x^2-8x+12=0a n dy^2-14 y+45=0`is`(4,7)``(7,4)``(9,4)``(4,9)`A. (4,7)B. (7,4)C. (9,4)D. (4,9)
Answer» Correct Answer - A

Discussion

382.	If a circle passes through the points (0, 0), (a, 0), and (0, b), then find the co-ordinates of its centre.(a) (-a/2, -b/2)(b) (a/2, -b/2)(c) (-a/2, b/2)(d) (a/2, b/2)
Answer» Correct option is (d) (a/2, b/2)

Discussion

383.	`/_ACB ` is inscribed in arc ACB of a circle with centre O . If `/_ACB= 65^(@)` , find m ( arc ACB ) .A. `65^(@)`B. `130^(@)`C. `295^(@)`D. `230^(@)`
Answer» Correct Answer - D

Discussion

384.	If two circles touch externally , how many common tangents can be drawn to them ?A. OneB. TwoC. ThreeD. Four
Answer» Correct Answer - C

Discussion

385.	The centres of those circles which touch the circle, `x^2+y^2-8x-8y-4=0`, externally and also touch thex-axis, lieon :(1) a circle.(2) an ellipse which is not acircle.(3) a hyperbola.(4) a parabola.A. a hyperbolaB. a parabolaC. a circleD. an ellipse which is not a circle
Answer» Correct Answer - B

Discussion

386.	The centres of those circles which touch the circle, `x^2+y^2-8x-8y-4=0`, externally and also touch thex-axis, lieon :(1) a circle.(2) an ellipse which is not acircle.(3) a hyperbola.(4) a parabola.A. a circleB. an ellipse which is not a circleC. a hyperbola0D. a parabola
Answer» Given equation ofcircle is `x^(2)+y^(2)-8x-8y-4=0`, whose centre is C(4, 4) and radius `=sqrt(4^(2)+4^(2)+4) = sqrt36 = 6` Let the centre of required circle be `C_(1)(x, y)`. Now, as it touch the X-axis, therefore its radius ` = \|y\|`. Also, it touch the circle `x^(2)+y^(2)-8x-8y-4=0 " therefore " "CC"_(1)=6+\|y\|` `rArr sqrt((x-4)^(2)+(y-4)^(2))=6+\|y\|` `rArr x^(2)+16-8x+y^(2)+16-8y=36+y^(2)+12\|y\|` `rArr x^(2)-8x-8y+32=36+12\|y\|` `rArr x^(2)-8x-8y-4=12\|y\|` Case I if `ygt0`, then we have `x^(2)-8x-8y-4=12y` `rArr x^(2)-8x-20y-4=0` `rArr (x-4)^(2)-20=20y` `rArr (x-4)^(2)=20(y+1)`, which is a parabola Case II if `ylt 0`, then we have `x^(2)-8x-8y-4=-12y` `rArr x^(2)-8x-8y-4+12y=0` `rArr x^(2)-8x+4y-4=0` `rArr x^(2)-8x-4=-4y` `rArr (x-4)^(2)=20-4y` `rArr(x-4)^(2)=-4(y-5)`, which is again a parabola .

Discussion

387.	Two circles with radii 3 cm and 2.5 cm touch each other externally then find the distance between their centre.
Answer» Distance between their centres = Sum of their radii …( By theorem of touching circles ) `= 3+ 2.5 ` `= 5.5 cm ` Distance between their centres is 5.5 cm .

Discussion

388.	Two circles of radii 5 . 5 cm and 3.3 cm respectively touch each other. What is the distance between their centres ?A. `4.4` cmB. `8.8 `cmC. `2.2` cmD. `8.8` or `2.2` cm
Answer» Correct Answer - B::C

Discussion

389.	If two circles have radii 10 cm and 15cm and they touch each other. Find the distance between their centres.
Answer» Correct Answer - Distance between their centres is 25 cm or 5 cm . Let the radii of the circles be `r_(1)` and `r_(2)` `r_(1) = 10cm` and `r_(2) = 15 cm ` By theorem of touching circles, the centres and point of contact are collinear . Here there can be two cases. Case 1`:` The circles touch each other externally Distance between their centres `= r_(1) + r_(2)` `= 10 + 15` `= 25 cm ` Case 2 `:` The circle touch each other internally Distance between their centres `= r_(2) - r_(1)` `= 15- 10 ` `= 5 cm `

Discussion

390.	Two circles of radii 5.5 cm and 3.3cm respectively touch each other. What is the distance between their centre ?A. 4.4 cmB. 8.8 cmC. 2.2 cmD. 8.8 or 2.2 cm
Answer» Correct Answer - D

Discussion

391.	In the figure, two circles with centres `A and B` and of radii `5cm and 3cm` touch each other intermally. If the perpendicular bisectors of segment AB meets the bigger circle in `P and Q.` Find the length of `PQ.`
Answer» Here, AM=5cm,BM=3cm. `thereforeAB=AM-BM=5-3=2cm` It is also given that PQ is the perpendicular bisector of AB. `therefore AC=CB=(1)/(2)AB=1cm` and `anglePCA=90^@` Now, in right triangle PCA, `PC^2=AP^2-AC^2` (by Pythagoras theorem) `=(5)^2-(1)^2=24` `therefore PC =sqrt(24)cm` `rArrPC=2(sqrt6)cm` `therefore PQ=2PC` (because perpendicular drawn from the centre to the chord bisects the chord `=2xx2sqrt(6)cm` `rArrPQ=4sqrt(6)cm`.

Discussion

392.	If the lines `3x-4y+4=0`and `6x-8y-7=0`are tangents to a circle, then find the radius of the circle.A. `(3)/(2)`B. 3C. `(5)/(2)`D. 5
Answer» Correct Answer - A

Discussion

393.	The length of common chord of two circles is 30 cm. if the diameters of circles are 50 cm and 34 cm, then find the distance between these centres.
Answer» Correct Answer - 28 cm

Discussion

394.	Every student in the group should do this activity. Draw a circle in your notebook. Draw any chord of that circle. Draw perpendicular to the chord through the centre of the circle. Measure the lengths of the two parts of the chord. Group leader should prepare a table as shown below and ask other students to write their observations in it. Write the property which you have observed.
Answer» On completing the above table, you will observe that the perpendicular drawn from the centre of a circle on its chord bisects the chord.

Discussion

395.	One of the diameters of the circle `x^2+y^2-12x+4y+6=0` is given byA. x+y=0B. x+3y=0C. x=yD. 3x+2y=0
Answer» Correct Answer - B

Discussion

396.	In the adjoining figure,`O` is the centre of two concentric circles. The chord AB of larger circle intersects the smaller circle at C and D. (i) Find `AC:BD`. (ii) If `AC=2cm`, then find the length of BD.
Answer» Correct Answer - (i) `1:1` (ii) 2 cm

Discussion

397.	In the adjoining figure, CD is a diameter of the circle with centre O. Diameter CD is perpendicular to chord AB at point E. Show that ∆ABC is an isosceles triangle.
Answer» Given: O is the centre of the circle. Diameter CD ⊥ chord AB, A-E-B To prove: ∆ABC is an isosceles triangle. Proof: diameter CD ⊥ chord AB [Given] ∴ seg OE ⊥ chord AB [C-O-E, O-E-D] ∴ seg AE ≅ seg BE ……(i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord] In ∆CEA and ∆CEB, ∠CEA ≅ ∠CEB [Each is of 90°] seg AE ≅ seg BE [From (i)] seg CE ≅ seg CE [Common side] ∴ ∆CEA ≅ ∆CEB [SAS test] ∴ seg AC ≅ seg BC [c. s. c. t.] ∴ ∆ABC is an isosceles triangle.

397.

In the adjoining figure, CD is a diameter of the circle with centre O. Diameter CD is perpendicular to chord AB at point E. Show that ∆ABC is an isosceles triangle.

Answer»

Given: O is the centre of the circle. Diameter CD ⊥ chord AB, A-E-B

To prove: ∆ABC is an isosceles triangle.

Proof:

diameter CD ⊥ chord AB [Given]

∴ seg OE ⊥ chord AB [C-O-E, O-E-D]

∴ seg AE ≅ seg BE ……(i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord]

In ∆CEA and ∆CEB, ∠CEA ≅ ∠CEB [Each is of 90°]

seg AE ≅ seg BE [From (i)]

seg CE ≅ seg CE [Common side]

∴ ∆CEA ≅ ∆CEB [SAS test]

∴ seg AC ≅ seg BC [c. s. c. t.]

∴ ∆ABC is an isosceles triangle.

Discussion

398.	Find the equation of the circle with centre `(2,2)` and passing through the point `(4,5)`.A. `x^(2)+y^(2)+4x+4y-5=0`B. `x^(2)+y^(2)-4x-4y-5=0`C. `x^(2)+y^(2)-4x-13=0`D. `x^(2)+y^(2)-4x-4y+5=0`
Answer» Correct Answer - B

Discussion

399.	In the adjoining figure, centre of two circles is O. Chord AB of bigger circle intersects the smaller circle in points P and Q. Show that AP = BQ.
Answer» Given: Two concentric circles having centre O. To prove: AP = BQ Construction: Draw seg OM ⊥ chord AB, A-M-B Proof: For smaller circle, seg OM ⊥ chord PQ [Construction, A-PM, M-Q-B] ∴ PM = MQ …..(i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord.] For bigger circle, seg OM ⊥ chord AB [Construction] ∴ AM = MB [Perpendicular drawn from the centre of the circle to the chord bisects the chord.] ∴ AP + PM = MQ + QB [A-P-M, M-Q-B] ∴ AP + MQ = MQ + QB [From (i)] ∴ AP = BQ

399.

In the adjoining figure, centre of two circles is O. Chord AB of bigger circle intersects the smaller circle in points P and Q. Show that AP = BQ.

Answer»

Given: Two concentric circles having centre O.

To prove: AP = BQ

Construction: Draw seg OM ⊥ chord AB, A-M-B

Proof:

For smaller circle,

seg OM ⊥ chord PQ [Construction, A-PM, M-Q-B]

∴ PM = MQ …..(i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]

For bigger circle, seg OM ⊥ chord AB [Construction]

∴ AM = MB [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]

∴ AP + PM = MQ + QB [A-P-M, M-Q-B]

∴ AP + MQ = MQ + QB [From (i)]

∴ AP = BQ

Discussion

400.	The equation of the circle passing through `(2,0)` and `(0,4)` and having minimum radius isA. `x^(2)+y^(2)=20`B. `x^(2)+y^(2)-2x-4y=0`C. `(x^(2)+y^(2)-4)+lamda(x^(2)+y^(2)-16)=0`D. None of the above
Answer» Correct Answer - B

Discussion

Explore topic-wise InterviewSolutions in .

In the figure, m ( arc PAQ ) =m (arc RBS ) `=130^(@)` chord PQ has length 10cm then what is the length of chord RS ? Why ?

In the adjoining figure , `O` is the centre of the circle. If `angleACB=25^@`, then find `angleAOB`.

In the adjoining figure, `O` is the centre of the circle. If `angle BOP=130^@`, then find the value of x.

`O` is the centre of a circle of diameter AB. If chord `AC=` chord BC, then find the value of `angleABC`.

In the adjoining figure, if `angleACB=40^@`, `angle DPB=120^@`, then find `angleCBD`.

In the adjoining figure, AOB is the diameter and `O` is the centre of the circle. If `angle BDC=60^@`, then find the value of `angleABC`.

Show that if two chords of a circle bisect one another they must bediameters.

In the adjoining figure, `O` is the centre of the circle. If `angle AOB= 150^@` and `angleBOC=100^@`, then find the value of `angleABC`.

In the adjoining figure, if ABCD is a cyclic quadrilateral, find the value of x.

In the adjoining figure, `O` is the centre of the circle and AC is its diameter. If `angleBAC=30^@`, then find `angleBOC`.

In fig. ∠ACB = 40°, find ∠OAB

In figure, if ∠OAB = 40°, then ∠ACB equals(a) 50°(b) 40°(c) 60°(d) 70°

In the adjoining figure, ABCD is a cyclic quadrilateral whose side AB is the diameter of the circle . If `angle ADC=140^(@)` , then find the value of `angleBAC`.

In the adjoining figure, `O` is the centre of the circle. If `angleBOD=50^@`, then find `angleBCD`. (ii) In the adjoining figure,`O` is the centre of the circle. If `angleABC=20^@`, then find `angleAOC`.

Number of circles passes through three collinear points(a) One(b) Two(c) Zero(d) Infinite

In the figure, two circles intersect each other at points A and B. AP and AQ are the diameters of these circels. Prove that PBQ is a straight line.

In the adjoining figure, ABCD is a cyclic quadrilateral. If `angleDBC=60^(@)` and `angleBAC=40^(@)` , then find the value of `angleBCD` .

In the adjoining , `O` is the centre of the circle. `ACB` is a segment. If `angle OAB=30^(@)`, then find the value of `angle ACB`.

Points A,B,C are on a circle, such that m( arc AB ) = m ( arc BC ) `=120^(@)` . No point, except point B, is common to the arcs. What type is the `Delta ABC `?A. Equilateral triangleB. Scalene triangleC. Right angled triangleD. Isosceles triangle

In the adjoining figure, AC is the diameter of the circle. If `angleBDC=115^(@)`, then find the value of `angle ACB`.

In a cyclic `square ` ABCD, twice the measure of `/_A` is thrice the measure of `/_C ` . Find the measure of `/_C `.A. `36^(@)`B. `72^(@)`C. `90^(@)`D. `108^(@)`

In the adjoining figure, `Delta ABC` is an isosceles triangle. Find the value of `angleBDC ` and `angleBEC`.

An angle of a cyclic trapezium is twice the other angle. Find the value of the smaller angle.

If O is the centre of a circle with radius r and AB is a chord of the circle at a distance `r/2` from O, then `/_BAO=`

Find the radius of the smalles circle which touches the straight line `3x-y=6`at `(-,-3)`and also touches the line `y=x`. Compute up to one place of decimal only.

`square` PQRS is cyclic .T is a point of ray SR such that S-R-T and `/_QRT 100^(@)` then what is the measure of `/_ SPQ` ? Why ?

In adjoining figure, PQ = QR . `angleP = 60^(@)` `:.` m (arc PR)= ..... A. `120^(@)`B. `60^(@)`C. `90^(@)`D. `240^(@)`

In the figure, O is the centre of circle `/_ QPR = 70^(@)` and m (arc PYR ) `= 160^(@)`, then find the value of each of the following `:` (a) m ( arc QXR ) (b) `/_QOR` (c ) `/_ PQR `

In the figure , `Delta QRS ` is an equilateral triangle. Prove that (1) arc RS `~= arc QS ~= ` arc QR (2) m (arc QRS ) `= 240^(@)`. Proof `:`

In the figure, P is the cnetre of the circle. If `/_ QPR 50^(@), /_ RPS = 60^(@) , /_ SPT = 100^(@)` then find (1) m(arc QRS) (ii) m ( arc QST ) (iii) m (arc RTS )

The centre of circleinscribed in a square formed by lines `x^2-8x+12=0a n dy^2-14 y+45=0`is`(4,7)``(7,4)``(9,4)``(4,9)`A. (4,7)B. (7,4)C. (9,4)D. (4,9)

If a circle passes through the points (0, 0), (a, 0), and (0, b), then find the co-ordinates of its centre.(a) (-a/2, -b/2)(b) (a/2, -b/2)(c) (-a/2, b/2)(d) (a/2, b/2)

`/_ACB ` is inscribed in arc ACB of a circle with centre O . If `/_ACB= 65^(@)` , find m ( arc ACB ) .A. `65^(@)`B. `130^(@)`C. `295^(@)`D. `230^(@)`

If two circles touch externally , how many common tangents can be drawn to them ?A. OneB. TwoC. ThreeD. Four

The centres of those circles which touch the circle, `x^2+y^2-8x-8y-4=0`, externally and also touch thex-axis, lieon :(1) a circle.(2) an ellipse which is not acircle.(3) a hyperbola.(4) a parabola.A. a hyperbolaB. a parabolaC. a circleD. an ellipse which is not a circle

The centres of those circles which touch the circle, `x^2+y^2-8x-8y-4=0`, externally and also touch thex-axis, lieon :(1) a circle.(2) an ellipse which is not acircle.(3) a hyperbola.(4) a parabola.A. a circleB. an ellipse which is not a circleC. a hyperbola0D. a parabola

Two circles with radii 3 cm and 2.5 cm touch each other externally then find the distance between their centre.

Two circles of radii 5 . 5 cm and 3.3 cm respectively touch each other. What is the distance between their centres ?A. `4.4` cmB. `8.8 `cmC. `2.2` cmD. `8.8` or `2.2` cm

If two circles have radii 10 cm and 15cm and they touch each other. Find the distance between their centres.

Two circles of radii 5.5 cm and 3.3cm respectively touch each other. What is the distance between their centre ?A. 4.4 cmB. 8.8 cmC. 2.2 cmD. 8.8 or 2.2 cm

In the figure, two circles with centres `A and B` and of radii `5cm and 3cm` touch each other intermally. If the perpendicular bisectors of segment AB meets the bigger circle in `P and Q.` Find the length of `PQ.`

If the lines `3x-4y+4=0`and `6x-8y-7=0`are tangents to a circle, then find the radius of the circle.A. `(3)/(2)`B. 3C. `(5)/(2)`D. 5

The length of common chord of two circles is 30 cm. if the diameters of circles are 50 cm and 34 cm, then find the distance between these centres.

One of the diameters of the circle `x^2+y^2-12x+4y+6=0` is given byA. x+y=0B. x+3y=0C. x=yD. 3x+2y=0

In the adjoining figure,`O` is the centre of two concentric circles. The chord AB of larger circle intersects the smaller circle at C and D. (i) Find `AC:BD`. (ii) If `AC=2cm`, then find the length of BD.

In the adjoining figure, CD is a diameter of the circle with centre O. Diameter CD is perpendicular to chord AB at point E. Show that ∆ABC is an isosceles triangle.

Find the equation of the circle with centre `(2,2)` and passing through the point `(4,5)`.A. `x^(2)+y^(2)+4x+4y-5=0`B. `x^(2)+y^(2)-4x-4y-5=0`C. `x^(2)+y^(2)-4x-13=0`D. `x^(2)+y^(2)-4x-4y+5=0`

In the adjoining figure, centre of two circles is O. Chord AB of bigger circle intersects the smaller circle in points P and Q. Show that AP = BQ.

The equation of the circle passing through `(2,0)` and `(0,4)` and having minimum radius isA. `x^(2)+y^(2)=20`B. `x^(2)+y^(2)-2x-4y=0`C. `(x^(2)+y^(2)-4)+lamda(x^(2)+y^(2)-16)=0`D. None of the above