Anúncio

Anúncio

Anúncio

Próximos SlideShares

Straight line and circle.pdf

Carregando em ... 3

28 de Mar de 2023•0 gostou## 0 gostaram

•4 visualizações## visualizações

Seja o primeiro a gostar disto

mostrar mais

Vistos totais

0

No Slideshare

0

De incorporações

0

Número de incorporações

0

Baixar para ler offline

Denunciar

Educação

Question bank on Circle & Straight line There are 115 questions in this question bank. Select the correct alternative : (Only one is correct) Q.14/circle Coordinates of the centre of the circle which bisects the circumferences of the circles x2 + y2 = 1 ; x2 + y2 + 2x – 3 = 0 and x2 + y2 + 2y – 3 = 0 is (A) (–1, –1) (B) (3, 3) (C) (2, 2) (D*) (– 2, – 2) Q.26/st.line One side of a square is inclined at an acute angle with the positive x-axis, and one of its extremities is at the origin. If the remaining three vertices of the square lie above the x-axis and the side of a square is 4, then the equation of the diagonal of the square which is not passing through the origin is (A) (cos + sin ) x + (cos – sin ) y = 4 (B) (cos + sin ) x – (cos – sin ) y = 4 (C*) (cos – sin ) x + (cos + sin ) y = 4 (D) (cos – sin ) x – (cos + sin ) y = 4 cos 2 4 sin 4 cos [Sol. slope of line AC = 4 cos 4 sin equation of line = sin sin cos sin y – sin = sin sin cos sin (x – 4cos ) y cos + y sin – 4 sin · cos = x sin – x cos – 4 sin2 – sin cos + 4 cos2 x (cos – sin ) + y (cos + sin ) = 4 (D) ] Q.310/circle The line 2x – y + 1 = 0 is tangent to the circle at the point (2, 5) and the centre of the circles lies on x – 2y = 4. The radius of the circle is (A*) 3 5 (B) 5 (C) 2 (D) 5 [Sol. 2x – y + 1 = 0 is tangent 1 slope of line OA = – 2 1 equation (y – 5) = – 2 (x – 2) 2y – 10 = – x + 2 x + 2y = 12 intersection with x – 2y = y will give coordinates of centre x + 2y = 12 – x – 2y = 4 + ——————— 4y = 8 y = 2 x – 4 = 4 x = 8 C = (8, 2) distance OA = = = = 3 ] Q.412/st.line Given the family of lines, a (2x + y + 4) + b (x 2y 3) = 0 . Among the lines of the family, the number of lines situated at a distance of from the point M(2, 3) is : (A) 0 (B*) 1 (C) 2 (D) [Hint: The point of intersection of the two lines are (–1, –2) Distance PM = 10 Hence the required line is one which passes through (–1, –2) and is | to P.M. B] Q.513/circle The co-ordinate of the point on the circle x² +y² 12x 4y+ 30 = 0, which is farthest from the origin are : (A*) (9 , 3) (B) (8 , 5) (C) (12 , 4) (D) none [Hint: radius = 36 + 4 – 30 = Ue section formula ] Q.628/st.line The area of triangle formed by the lines x + y – 3 = 0 , x – 3y + 9 = 0 and 3x – 2y + 1= 0 16 (A) 7 10 sq. units (B*) 7 sq. units (C) 4 sq. units (D) 9 sq. units Q.738/circle The number of common tangent(s) to the circles x² + y² + 2x+ 8y 23 = 0 and x² + y² 4x 10y+ 19 = 0 is : (A) 1 (B) 2 (C*) 3 (D) 4 Q.830/st.line The four points whose coordinates are (2, 1), (1, 4), (4, 5), (5, 2) form : (A) a rectangle which is not a square (B) a trapezium which is not a parallelogram (C*) a square (D) a rhombus which is not a square. [Sol. A (2, 1), B (1, 4), C(4, 5), D(5, 2) AB = BC = CD = DA = also AC BD i.e. mAC x mBD = -1 and AC = BD ] Q.941/circle From the point A(0 , 3) on th

STUDY INNOVATIONSSeguir

Educator em Study InnovationsAnúncio

- [Hint: Compute chord ofconstant of(–2, 0) and verifyeach alternative w.r.t. the and C.O.C. ] Q.121158/st.line Straight lines 2x + y= 5 and x 2y= 3 intersect at the pointA. Points B and C are chosen on these two lines such that AB =AC . Then the equation ofa line BC passing through the point (2, 3) is (A*) 3x y 3 = 0 (B*) x + 3y 11 = 0 (C) 3x + y 9 = 0 (D) x 3y + 7 = 0 [Hint: Note that the lines are perpendicular . Find the equation ofthe lines through (2, 3) and parallel to the bisectors of the given lines, the slopes of the bisectors being 1/3 & 3 ] Q.122144/circle The centre(s) of the circle(s) passing through the points (0, 0) , (1, 0) and touching the circle x2 + y2 = 9 is/are : (A) 3 2 1 2 , (B) 1 2 3 2 , (C*) 1 2 21 2 , / (D*) 1 2 21 2 , / [Hint: consider familyof 's through(0, 0) and (1, 0) x(x – 1) + y2 + y = 0 touches x2 + y2 = 9 common chord = – x + hy + 9 = 0 ....(1) perpendicular from(0, 0) on (1) is equal to 3. 2 1 9 = 3 2 = 8 = ± 2 2 circle x (x – 1) + y2 + 2 2 y centre 2 1 2 , 2 1 ] Q.123172/st.line The x co-ordinates of the vertices of a square of unit area are the roots of the equation x2 3x + 2 = 0 and the y co-ordinates of the vertices are the roots of the equation y2 3y+ 2 = 0 then the possible vertices of the square is/are : (A*) (1, 1), (2, 1), (2, 2), (1, 2) (B*) ( 1, 1), ( 2, 1), ( 2, 2), ( 1, 2) (C) (2, 1), (1, 1), (1, 2), (2, 2) (D) ( 2, 1), ( 1, 1), ( 1, 2), ( 2, 2) Q.124146/circle Acircle passes through the point 3 7 2 , and touches the line pair x2 y2 2x + 1 = 0. The co-ordinates ofthe centre ofthe circle are : (A*) (4, 0) (B) (5, 0) (C*) (6, 0) (D) (0, 4) Q.125180/st.line P(x, y) moves such that the area of the triangle formed by P, Q (a , 2a) and R (a, 2a) is equalto the area ofthe triangle formed by P, S (a, 2a) & T (2a, 3a). The locus of 'P' is a straight line givenby: (A*) 3x y = a (B*) 5x 3y + a = 0 (C) y = 2ax (D) 2y = ax
- Q.110178/circle In the xyplane, the segment with end points (3, 8) and (–5, 2) is the diameter ofthe circle. The point (k, 10) lies on the circle for (A) no value ofk (B*) exactlyone integralk (C) exaclyone nonintegralk (D) two realvalues ofk [Hint: k = – 1 ] Q.111199/st.line Let A (3, 2) and B (5, 1). ABP is an equilateral triangle is constructed on the side ofAB remote fromtheoriginthenthe orthocentre oftriangleABPis (A) 4 1 2 3 3 2 3 , (B) 4 1 2 3 3 2 3 , (C) 4 1 6 3 3 2 1 3 3 , (D*) 4 1 6 3 3 2 1 3 3 , [Sol. mAB = –1/2 mPM = 2 h = a 3 2 now a = 5 = AB ; h = 15 2 so point P x y cos = y 3 2 / sin = h P : x = 4 + 3 2 , y = 3 2 + 3 so orthocentre/centroid is x y 1 1 3 3 , = 4 + 3 6 , 3 2 + 3 3 ] Q.112200/st.line The vertexofa right angleofa right angledtriangle lies onthestraight line 2x+y– 10 =0 andthe two other vertices, at points (2, –3) and (4, 1) then the area oftriangle insq. units is (A) 10 (B*) 3 (C) 33 5 (D) 11 [Sol. M1M2 = –1 9 2 4 13 2 2 a a x a a = –1 117 – 26a – 18a + 4a2 = –(a2 – 6a + 8) 5a2 – 50a + 125 = 0 a = 5 so B is (5, 0) so area = 2 1 AB × AC = 2 1 2 × 3 2 = 3 ]
- Q.99188/st.line The coordinates of three points A(4, 0) ; B(2, 1) and C(3, 1) determine the vertices of an equilateraltrapeziumABCD . The coordinates ofthe vertex D are : (A) (6, 0) (B) ( 3, 0) (C) ( 5, 0) (D*) (9, 0) [Hint: Equilateralmeanisosceles trapezium] Q.100168/circle Tangents are drawn from anypoint on the circle x2 + y2 = R2 to the circle x2 + y2 = r2. Ifthe line joining the pointsofintersectionofthesetangents withthefirst circlealso touchthe second,thenRequals (A) 2 r (B*) 2r (C) 2 2 3 r (D) 4 3 5 r Q.101192/st.line The image ofthe pair oflines represented by ax2 + 2h xy + by2 = 0 bythe line mirror y = 0 is (A) ax2 2h xy by2 = 0 (B) bx2 2h xy + ay2 = 0 (C) bx2 + 2h xy + ay2 = 0 (D*) ax2 2h xy + by2 = 0 [Hint : m1 m1 & m2 m2 equation is (y + m1 x) (y + m2 x) = 0 where m1 + m2 = 2h b & m1 m2 = a b ] Q.102169/circle Pair oftangents are drawn fromevery point on the line 3x + 4y = 12 on the circle x2 + y2 = 4. Their variable chordofcontact always passes through a fixed point whose co-ordinates are (A) 4 3 , 3 4 (B) 4 3 , 4 3 (C) (1, 1) (D*) 3 4 , 1 Q.103193/st.line The set ofvalues of 'b' for whichthe originand the point (1, 1) lieonthe same sideofthe straight line, a2x + a by + 1 = 0 a R, b > 0 are : (A) b (2, 4) (B*) b (0, 2) (C) b [0, 2] (D) (2, ) [Hint: a2 + ab + 1 > 0 a R D < 0 ] Q.104170/circle The equation of the circle symmetric to the circle x2 + y2 – 2x – 4y + 4 = 0 about the line x – y = 3 is (A*) x2 + y2 – 10x + 4y + 28 = 0 (B) x2 + y2 + 6x + 8 = 0 (C) x2 + y2 – 14x – 2y + 49 = 0 (D) x2 + y2 + 8x + 2y + 16 = 0 [Hint: Circle of the centre is (1, 2) and r = 1 image of (1, 2) in x – y – 3 = 0 1 1 a 2 b a + b =3 ....(1) also 2 2 b , 2 1 a lies on x – y = 3 (a + 1 ) – ( b + 2) = 6 a – b = 7 ....(2) from (1) and (2) a = 5 and b = – 2 Hence the required circle has the centre (5, – 2) and r = 1 (x – 5)2 + (y + 2)2 = 1 x2 + y2 – 10x + 4y + 28 = 0 (A) ]
- Q.91178/st.line Ifin triangleABC, A (1, 10) , circumcentre 1 3 2 3 , and orthocentre 11 3 4 3 , then the co-ordinates ofmid-point ofside opposite toAis : (A*) (1, 11/3) (B) (1, 5) (C) (1, 3) (D) (1, 6) Q.92159/circle Let x & y be the real numbers satisfying the equation x2 4x + y2 + 3 = 0. If the maximum and minimumvalues ofx2 + y2 are M & mrespectively, then the numericalvalue ofM mis : (A) 2 (B*) 8 (C) 15 (D) noneofthese Q.93181/st.line If the straight lines , ax + amy + 1 = 0 , bx + (m + 1) by + 1 = 0 and cx + (m + 2)cy + 1 = 0, m 0 are concurrent then a, b, c are in : (A) A.P. only for m = 1 (B) A.P. for all m (C) G.P. for all m (D*) H.P. for all m. [Sol. D = 1 ) 2 m ( c c 1 ) 1 m ( b b 1 am a = 0 ; 1 c 2 c 1 b b 1 0 a = 0 a(b – 2c) +(2bc – bc) = 0 ab – 2ac + bc = 0 b(a + c) = 2ac b = c a ac 2 a, b, c are in H.P. ] Q.94160/circle Aline meets the co-ordinate axes inA&B.Acircle is circumscribedabout the triangle OAB. Ifd1 &d2 are thedistances ofthe tangent to the circle at the originO fromthepointsAandB respectively, the diameter ofthecircle is : (A) 2 2 1 2 d d (B) d d 1 2 2 2 (C*) d1 + d2 (D) d d d d 1 2 1 2 Q.95182/st.line If x1 , y1 are the roots of x2 + 8 x 20 = 0, x2 , y2 are the roots of 4x2 + 32x 57 = 0 and x3 , y3 are the roots of 9 x2 + 72 x 112 = 0, then the points, (x1 , y1) , (x2 , y2) & (x3 , y3) (A*) are collinear (B) formanequilateraltriangle (C) forma right angled isosceles triangle (D) are concyclic [Sol. Given x1 + y1 = -8 y1 = -8-x1 ; x2 + y2 = -8 y2 = -8 - x2 and x3 + y3 = -8 y3 = -8 -x3 now check 1 x 8 x 1 x 8 x 1 x 8 x 3 3 2 2 1 1 value ofthis determinant comes out to be zero. ] Q.96164/circle Two concentric circles are such that the smaller divides the largerinto two regions ofequalarea. If the radius ofthesmaller circle is 2, then the lengthofthe tangent fromanypoint 'P'onthe larger circle to the smaller circle is : (A) 1 (B) 2 (C*) 2 (D) none [Sol. r1 2 = r2 2 r1 2 2 r1 2 = r2 2 r2 = 2 r1 Note P lies on the director circle of radius r1 L = r1 = 2 cm ]
- Q.82148/circle The circle passing through the distinct points (1, t) , (t, 1) & (t, t) for all values of 't ' , passes throughthepoint : (A) ( 1, 1) (B) ( 1, 1) (C) (1, 1) (D*) (1, 1) [Sol. Equation of circle is x2 + y2 + 2gx +2fy + c = 0 (1, t) 1 + t2 + 2g + 2ft + c = 0 (t, t) t2 + t2 + 2gt + 2ft + c = 0 (t, 1) 1 + t2 + 2gt + 2f + c = 0 subtract 1 + 2g – t2 – 2gt = 0 1 – t2 + 2g(1 – t) = 0 (1 – t)(1 + t + 2g) = 0 t = 1 one point (t, t) passes through(1, 1) ] Q.83173/st.line In a triangle ABC, side AB has the equation 2x + 3y = 29 and the sideAC has the equation, x + 2y = 16 . If the midpoint of BC is (5, 6) then the equation ofBC is : (A) x y = 1 (B) 5 x 2 y = 13 (C*) x + y = 11 (D) 3 x 4 y = 9 Q.84149/circle Ifa circle ofconstant radius 3k passes throughthe origin 'O' and meets co-ordinate axes atAand B thenthe locus ofthecentroid ofthe triangleOAB is (A*) x2 + y2 = (2k)2 (B) x2 + y2 = (3k)2 (C) x2 + y2 = (4k)2 (D) x2 + y2 = (6k)2 [Sol. Given r = 3k and P is centroid g2 + f2 = (3k)2 ....(1) because c = 0 3h1 = a and 3k1 = b also g 2 = a 2g = a and 2f = b 2g = 2 h 3 1 and 2f = 2 k 3 1 substitute in (1) 4 k 9 4 h 9 2 1 2 1 = 9k2 x2 + y2 = (2k)2 ] Q.85174/st.line The circumcentre of the triangle formed bythe lines, xy+ 2x+ 2y+ 4 = 0 and x + y+ 2 = 0 is (A) ( 2, 2) (B*) ( 1, 1) (C) (0, 0) (D) ( 1, 2) [Sol. ABC isright angle circumcentre is mid pt. ofBC i.e. (-1,-1)Ans. ] Q.86153/circle The locus ofthe midpoints ofthe chords ofthe circle x2 + y2 2x 4y 11 = 0 which subtend 600 at the centre is (A) x2 + y2 4x 2y 7 = 0 (B) x2 + y2 + 4x + 2y 7 = 0 (C*) x2 + y2 2x 4y 7 = 0 (D) x2 + y2 + 2x + 4y + 7 = 0 [Sol. (h, k) are mid points ofchord 4 ) k 2 ( ) h 1 ( 2 2 = cos 30° = (1 – h)2 + (2 – k)2 = 2 4 2 3 h2 + 1 – 2h + k2 + 4 – 4k = 12 hence locus is x2 + y2 – 2x – 4y – 7 = 0 ]
- p = 4 3 2 2 = 7 & c3M = 7 4 CM = 7 4 7 = 3 7 4 A = 3 7 4 . 3 2 . 1 2 = 9 7 16 ] Q.73155/st.line Let the co-ordinates of the two points A& B be (1, 2) and (7, 5) respectively. The line AB is rotated through 45º in anticlockwise directionabout thepoint oftrisectionofAB whichis nearerto B. The equationofthe line innew positionis : (A) 2x y 6 = 0 (B) x y 1 = 0 (C*) 3x y 11 = 0 (D) none of these [Hint : point oftrisectionis (5, 4) .After rotation the slopeofthe line tan 1 = 2 2 tan 45 tan 1 tan 45 tan = 2 1 2 1 1 1 = 3 C ] Q.74132/circle Apair oftangents are drawnto a unit circle with centre at the origin and these tangents intersect at A enclosing anangle of60°. The area enclosed bythese tangents and the arc ofthe circle is (A) 3 2 – 6 (B*) 3 – 3 (C) 3 – 6 3 (D) 6 1 3 [Hint: r = 1 ; L = 3 quad = 3 sector = 2 1 · 1 · 3 2 = 3 shaded region = 3 – 3 Ans. ] Q.75161/st.line The true set ofreal values of for which the point P with co-ordinate (, 2) does not lie inside the triangle formed by the lines, x y = 0 ; x + y 2 = 0 & x + 3 = 0 is : (A) (, 2] (B) [0, ] (C) [ 2, 0] (D*) (, 2] [0, ] [Hint : Point lies on the parabola y = x2 . Solve y = x2 with the line x + y = 2 ] Q.76138/circle If the line x cos + y sin = 2 is the equation of a transverse common tangent to the circles x2 + y2 = 4 and x2 + y2 6 3 x 6y + 20 = 0, then the value of is : (A) 5/6 (B) 2/3 (C) /3 (D*) /6 [Sol. C1C2 = r1 + r2 C1 = (0, 0) ; C2 = ( 3 3 , 3) & r1 = 2, r2 = 4 circle toucheachother T.C.T = 3 x + y – 4 = 0 comparing with x cos + y sin = 2 = 6 ]
- Q.64124/circle If two chords, each bisected by the xaxis canbe drawn to the circle, 2 (x2 + y2) 2ax by = 0 (a 0 , b 0) from the point (a, b/2) then : (A) a2 > 8b2 (B) b2 > 2a2 (C*) a2 > 2b2 (D) a2 = 2b2 [Hint: (h, b/2) must lie onthe given circle 2h2 2ah + b2 = 0 For two distinct values of origin h D > 0 a2 > 2b2 C ] Q.65146/st.line Ifthe line y= mx bisects the angle between the lines ax2 + 2h xy + by2 = 0 then m is a root of the quadraticequation: (A*) hx2 + (a b) x h = 0 (B) x2 + h(a b) x 1 = 0 (C) (a b) x2 + hx (a b) = 0 (D) (a b) x2 hx (a b) = 0 [Hint : Equation of the angle bisector x y a b 2 2 = xy h ; Now put y = mx ] Q.66125/circle Tangents aredrawnto a unit circle withcentre at the originfromeachpoint onthe line 2x + y= 4. Then the equation to the locus ofthe middle point ofthe chord ofcontact is (A) 2 (x2 + y2) = x + y (B) 2 (x2 + y2) = x + 2y (C*) 4 (x2 + y2) = 2x + y (D) none [Sol. (x1, y1) lies on 2x + y = 4 2x1 + y1 = 4 ....(1) chord ofcontact w.r.t. (x1, y1) xx1 + yy1 = 1 also equation ofchord whose mid point is (h, k) h2 + k2 = hx + ky 2 2 1 1 k h 1 k y h x x1 = 2 2 k h h ; y1 = 2 2 k h k substitute in(1) 2 · 2 2 k h h + 2 2 k h k = 4 locus = 4(x2 + y2) = 2x + y ] Q.67149/st.line An equilateraltriangle has each of its sides of length 6 cm. If(x1, y1) ; (x2, y2) & (x3, y3) are its vertices thenthevalue ofthe determinant, 2 3 3 2 2 1 1 1 y x 1 y x 1 y x is equal to : (A) 192 (B) 243 (C) 486 (D*) 972 Q.68126/circle Two circles whose radiiare equalto 4 and 8 intersect at right angles. The lengthoftheir common chord is (A*) 16 5 (B) 8 (C) 4 6 (D) 8 5 5
- Q.54116/circle The chord ofcontact of the tangents drawn from a point on the circle, x2 + y2 = a2 to the circle x2 + y2 = b2 touches the circle x2 + y2 = c2 then a, b, c are in : (A) A.P. (B*) G.P. (C) H.P. (D) A.G.P. [Hint: equation ofchordofcontactAB x a cos + y a sin = b2 .....(1) this is tangent to the circle x2 + y2 = c2 perpendicular from(0, 0) on (1) is equalto c 2 2 2 2 2 sin b cos a b = c b2 = ac ] Q.55106/st.line Alight beamemanating fromthe pointA(3, 10) reflects fromthe line 2x+ y- 6 = 0and thenpasses throughthe point B(5, 6) . The equationofthe incident and reflected beams are respectively : (A*) 4 x 3 y + 18 = 0 & y = 6 (B) x 2 y + 8 = 0 & x = 5 (C) x + 2 y 8 = 0 & y = 6 (D) none of these [Hint : slope ofAB = 2 = slope ofgiven line PAB is isosceles . Equation ofline MPis x 2y + 12 = 0 . Solving it with 2x + y 6 = 0 point P(0, 6) . Equation ofAP is 4x 3y + 18 = 0 and equation of BP is y 6 = 0 ] [ M.E. with Q. No. 91 ] Q.56118/circle If the two circles, x2 + y2 + 2 g1x + 2 f1y = 0 & x2 + y2 + 2 g2x + 2 f2y = 0 touch each then: (A) f1 g1 = f2 g2 (B*) f g 1 1 = f g 2 2 (C) f1 f2 = g1 g2 (D) none [Hint: circles can touchonlyat (0, 0) (–g, –f1); (0, 0) and (–g2, –f2) collinear f g f g 1 1 2 2 ] Q.57109/st.line If P 1 x p p , ; Q = 1 x q q , ; R = 1 x r r , where xk 0, denotes the kth term of an H.P. . for k N, then : (A) Ar. ( PQR) = p q r p q q r r p 2 2 2 2 2 2 2 ( ) ( ) ( ) (B) PQR is a right angled triangle (C*) the points P, Q, R are collinear (D) none [Hint : 1 xp = a + (p 1) d ; 1 xq = a + (q 1) d ; 1 xr = a + (r 1) d 1 r d ) 1 r ( a 1 q d ) 1 q ( a 1 p d ) 1 p ( a = 0] Q.58119/circle Tangents are drawn to the circle x2 + y2 = 1 at the points where it is met by the circles, x2 + y2 ( + 6)x + (8 2) y 3 = 0 . being the variable . The locus ofthe point ofintersection of these tangentsis : (A*) 2x y+ 10 = 0 (B) x + 2y 10 = 0 (C) x 2y+ 10 = 0 (D) 2x + y 10 = 0 [Hint: compare chord ofcontact ofthe pair oftangents from(x1, y1) to thecircle x2 + y2 = 1 withthe common chord betweenthe two circles and eliminate ] [Sol. Locus ofpoint ofintersection oftangents chord of contact of (x1, y1) w.r.t. x2 + y2 = 1 is xx1 + yy1 = 1 (AB) .....(1)
- [Hint: equation of the two circles be (x r)2 + (y r)2 = r2 i.e. x2 + y2 2rx 2ry+ r2 = 0 where r = r1 & r2. Condition of orthogonality gives 2 r1r2 + 2 r1r2 = r1 2 + r2 2 4 r1r2 = r1 2 + r2 2. Circle passes through (a, b) a2 + b2 2ra 2rb + r2 = 0 i.e. r2 2r (a + b) + a2 + b2 = 0 r1 + r2 = 2 (a + b) and r1 r2 = a2 + b2 ] Q.4785/st.line Ifthe verticesPand Q ofa triangle PQR are given by(2, 5)and (4, –11) respectively, and the point R moves along the line N: 9x+ 7y+ 4= 0, thenthelocus ofthe centroidofthe triangle PQR is a straight line parallelto (A) PQ (B) QR (C) RP (D*) N Sol. R (x, y) lies on 9x + 7y + 4 =0 7 a 9 4 , a R , centroid of PQR = (h, k) 3 a 6 3 a 4 2 h ....(1) 3 7 a 9 46 3 7 ) a 9 4 ( 11 5 k ....(2) from(1) &(2) we get equatingx 0 46 54 21 27 9 ) 46 21 ( 6 3 k h k h or locus is 9x + 7y - 8/3 = 0 this line is to N ] Q.48111/circle The range ofvalues of 'a' suchthat the angle between the pair oftangents drawnfromthe point (a, 0) to the circle x2 + y2 = 1 satisfies 2 < < is : (A) (1, 2) (B) 1 2 , (C) 2 1 , (D*) 2 1 , 1 2 , [Hint: Use the concept ofdirector circle ] Q.4989/st.line The pointsA(a, 0), B(0, b), C(c, 0) & D(0, d) are suchthat ac = bd & a, b, c, d are allnonzero. The the points : (A) forma parallelogram (B) do not lie on a circle (C) forma trapezium (D*) are concyclic [Hint : a/d = b/c cyclic ] Q.50112/circle If(,) is a point on the circle whose centre is on the x -axis and which touches the line x + y= 0 at (2, –2), thenthe greatest value ofis (A) 4 – 2 (B) 6 (C*) 4 + 2 2 (D) 4 + 2
- Q.3566/st.line Astick oflength10 unitsrests against the floor&a wallofa room. Ifthe stick begins to slide onthe floor thenthelocus ofits middlepoint is : (A) x2 + y2 = 2.5 (B*) x2 + y2 = 25 (C) x2 + y2 = 100 (D) none [Sol. given (a2 + b2) = 100 ---(1) p (h, k) be the middle pt then h = a/2, k = b/2 a = 2h b = 2k from(1) 4h2 + 4k2 = 100 h2 + k2 =25 locus x2 + y2 = 25 ] Q.3680/circle The locus of the mid points ofthe chords ofthe circle x²+y²+4x6y12 = 0 which subtend an angle of 3 radiansat its circumference is : (A) (x 2)² + (y + 3)² = 6.25 (B*) (x + 2)² + (y3)² = 6.25 (C) (x + 2)² + (y 3)² = 18.75 (D) (x + 2)² + (y + 3)² = 18.75 [Hint : radius = 5 p = 5 cos 60º = 2.5 locus is (h + 2)2 + (k 3)2 = 6.25 ] Q.3772/st.line Througha givenpoint P(a, b)astraight line is drawnto meet the axesat Q &R.Ifthe parallelogram OQSR is completed then the equationofthe locus ofS is (given 'O' isthe origin) : (A*) a x + b y = 1 (B) a y + b x = 1 (C) a x + b y = 2 (D) a y + b x = 2 Q.3892/circle The points (x1, y1) , (x2, y2), (x1, y2) & (x2, y1) are always : (A) collinear (B*) concyclic (C) vertices of a square (D) vertices ofa rhombus [Hint : All the points lie on the circle (x x1) (x x2) + (y y1) (y y2) = 0 ] Q.39202/st.line The numberofpossible straight lines, passing through(2, 3) and forming atriangle withcoordinate axes, whose area is 12 sq. units , is (A) one (B)two (C*) three (D)four [Sol equation ofanyline through (2, 3) is y -3 = m(x - 2) y = mx - 2m + 3 with the help ofthe fig. area of OAB = 12 ie. 12 ) 2 3 ( 3 2 2 1 m m m B A O (0,0) .(2,3) (0, 3–2m) taking + sign me get (2m+3)2 = 0 this gives one value ofm= -3/2 takingnegativesignwe get 4m2 - 36m + 9 = 0 (D > 0) quadratic inmgives 2 values ofm 3 st. lines are possible. ] Q.4073/st.line Two mutuallyperpendicular straight lines throughthe origin froman isosceles trianglewiththe line 2x + y= 5 . Thenthe area ofthe triangle is : (A*) 5 (B) 3 (C) 5/2 (D) 1
- Q.2367/circle The equation of the circle having the lines y2 2y+4x2xy= 0 as its normals & passing through the point (2, 1) is : (A*) x2 + y2 2x 4y + 3 = 0 (B) x2 + y2 2x + 4y 5 = 0 (C) x2 + y2 + 2x + 4y 13 = 0 (D) none [Sol. normals are (y– 2)(y– 2x) = 0 should pass throughcentre (a, b) 2a = b and b = 2 a = 1 and b = 2 equation x2 + y2 – 2x – 4y + c = 0 ] Q.2448/st.line If P = (1, 0) ; Q = (1, 0) & R = (2, 0) are three given points, then the locus of the points S satisfying the relation, SQ2 + SR2 = 2SP2 is : (A) a straight line parallelto xaxis (B) a circle passing throughthe origin (C) a circle with the centre at the origin (D*) a straight line parallelto yaxis . Q.2570/circle Ifa circle passes through the point (a, b) & cuts the circle x²+ y² = K² orthogonally, then the equation of the locus of its centre is : (A*) 2ax + 2by (a² + b² + K²) = 0 (B) 2ax + 2by (a² b² + K²) = 0 (C) x² + y² 3ax 4by+ (a² + b² K²) = 0 (D) x² + y² 2ax 3by+ (a² b² K²) = 0 Q.2650/st.line The coordinates ofthe orthocentre ofthe triangle bounded bythe lines, 4x7y+10 = 0;x + y=5 and 7x + 4y = 15 is : (A) (2, 1) (B) ( 1, 2) (C*) (1, 2) (D) (1, 2) Q.2774/circle The distance between the chords of contact of tangents to the circle ; x2+y2 +2gx+2fy+ c=0 from the origin & the point (g, f) is : (A) g f 2 2 (B) g f c 2 2 2 (C*) g f c g f 2 2 2 2 2 (D) g f c g f 2 2 2 2 2 [Sol. Equation ofchord or contact are gx + fy + c = 0 ....(1) & 2gx + 2fy + 2 c f g 2 2 = 0 ....(2) These lines are parallel hence distance = 2 2 2 2 f g 2 c f g c ] Q.2852/st.line The equation ofthe pair ofbisectors ofthe angles between two straight lines is, 12x2 7xy 12y2 = 0 . If the equation of one line is 2y x = 0 then the equation of the other line is : (A*) 41x 38y = 0 (B) 38x 41y = 0 (C) 38x + 41y = 0 (D) 41x + 38y = 0 [Hint : (2y x) (y mx) = mx2 xy (2m + 1) + 2y2 = 0 the equation to the pair of bisectors are : x y m 2 2 2 = 2 2 1 xy m 12x2 7xy 12y2 m = 41/38 ] Q.2976/circle The points A(a, 0) , B (0 , b), C (c, 0) & D(0 , d) are such that ac = bd & a, b, c, d are all non-zero. Thenthepoints : (A) forma parallelogram (B) do not lie on a circle (C) forma trapezium (D*) are concyclic
- Q.1233/st.line IfA& B are the points (3, 4) and (2, 1), then the coordinates ofthe point C onAB produced such that AC = 2BC are : (A) (2, 4) (B) (3, 7) (C*) (7, 2) (D) 1 2 5 2 , Q.1352/circle The locus ofthe mid points of the chords ofthe circle x2 + y2 axby= 0 which subtend a right angle at a 2 b 2 , is : (A) ax + by = 0 (B) ax + by = a2 + b2 (C*) x2 + y2 ax by+ 8 b a 2 2 = 0 (D) x2 + y2 ax by 8 b a 2 2 = 0 [Sol. r = 4 b 4 a 2 2 = 2 b a 2 2 sin 45° = 2 b a 2 b k 2 a h 2 2 2 2 2 2 2 2 b a 4 ) b k 2 ( 4 ) a h 2 ( 4 2 1 simplify to get locus x2 + y2 – ax – by – 8 b a 2 2 = 0 ] Q.1434/st.line The base BCofa triangleABC is bisected at the point (p, q) and the equation to the sideAB &AC are px + qy = 1 & qx + py = 1 . The equation ofthe median throughAis : (A) (p 2q) x + (q 2p) y + 1 = 0 (B) (p + q) (x + y) 2 = 0 (C*) (2pq 1) (px + qy 1) = (p2 + q2 1) (qx + py 1) (D) none [Sol. EquationthroughAis (px + qy -1) + (qx + py - 1) = 0 .....(1) ifthis representAB then, point D (p, q) satisfies (1) [pp + qq – 1] + [pq + pq – 1] = 0 1 q p ) 1 pq 2 ( 2 2 ] Q.1555/circle From (3, 4) chords are drawn to the circle x² + y²4x = 0 . The locus of the mid points of the chords is : (A*) x² + y² 5x 4y + 6 = 0 (B) x² + y² + 5x 4y + 6 = 0 (C) x² + y² 5x + 4y + 6 = 0 (D) x² + y² 5x 4y 6 = 0 [Hint: Arc ofthe with OP as diameter intercepted bythe givencircle. ]
- Question bank on Circle & Straight line There are 115 questions in this question bank. Select the correct alternative : (Only one is correct) Q.14/circle Coordinates ofthe centre ofthe circle which bisects the circumferences ofthe circles x2 + y2 = 1 ; x2 + y2 + 2x – 3 = 0 and x2 + y2 + 2y – 3 = 0 is (A) (–1, –1) (B) (3, 3) (C) (2, 2) (D*) (– 2, – 2) Q.26/st.line One sideofa square is inclined at an acute angle withthe positive x-axis, and one ofits extremities is at theorigin. Ifthe remaining three vertices ofthe square lie above the x-axis and the side ofa square is 4, thenthe equation ofthe diagonalofthe square whichis not passing throughthe origin is (A) (cos + sin ) x + (cos – sin ) y = 4 (B) (cos + sin ) x – (cos – sin ) y = 4 (C*) (cos – sin ) x + (cos + sin ) y = 4 (D) (cos – sin ) x – (cos + sin ) y = 4 cos 2 [Sol. slope of lineAC = sin 4 cos 4 cos 4 sin 4 equationofline = sin cos sin sin y – sin = sin cos sin sin (x – 4cos ) y cos + y sin – 4 sin · cos = x sin – x cos – 4 sin2 – sin cos + 4 cos2 x (cos – sin ) + y (cos + sin ) = 4 (D) ] Q.310/circle The line 2x – y+ 1 = 0 is tangent to the circle at the point (2, 5) and the centre ofthe circleslies on x – 2y= 4. The radius ofthe circle is (A*) 5 3 (B) 3 5 (C) 5 2 (D) 2 5 [Sol. 2x – y + 1 = 0 is tangent slope of line OA = – 2 1 equation (y – 5) = – 2 1 (x – 2) 2y – 10 = – x + 2 x + 2y = 12 intersectionwithx – 2y= ywillgive coordinatesofcentre x + 2y = 12 – x – 2y = 4 + ——————— 4y = 8 y = 2 x – 4 = 4 x = 8 C = (8, 2) distance OA = 2 ) 5 2 ( ) 2 8 ( = 9 36 = 45 = 5 3 ]

Anúncio