Saved Bookmarks
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
In the figure, two circles intersect at points M and N. Secants drawn through M and N intersect the circles at points R,S and P,Q respectively. Prove that `:` seg SQ `||` seg RP. |
|
Answer» Draw seg MN `square ` MRPN is cyclic and `/_ MNQ ` is its exterior angle. `/_ MNQ = /_ MRP ` ….(1) …...(Corollary of cyclic quadrilateral theorem ) `square ` MNQS is cyclic `:. /_ MNQ + /_ MSQ = 180^(@0` …(Theorem of cyclic quadrilateral ) `:. /_ MRP + /_MSQ = 180^(@)` ....[From (1) ] `:. /_SRP + /_ RSQ = 180^(@)` .....(R-M-S) `:.` seg `SQ || ` seg `RP ` ....(interior angles test for parallel lines ) |
|
| 52. |
In Fig., if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to:(A) 2 cm(B) 3 cm(C) 4 cm(D) 5 cm |
|
Answer» (A) 2 cm Explanation: Given: Radius of the circle = r = AO = 5 cm Length of chord AB = 8 cm Since the line drawn through the center of a circle to bisect a chord is perpendicular to the chord, therefore AOC is a right angled triangle with C as the bisector of AB. ∴ AC = ½ (AB) = 8/2 = 4 cm In right angled triangle AOC, by Pythagoras theorem, we have: (AO)2 = (OC)2 + (AC)2 ⇒ (5)2 = (OC)2 + (4)2 ⇒ (OC)2 = (5)2 – (4)2 ⇒ (OC)2 = 25 – 16 ⇒ (OC)2 = 9 Take square root on both sides: ⇒ (OC) = 3 ∴ The distance of AC from the center of the circle is 3 cm. Now, OD is the radius of the circle, ∴ OD = 5 cm CD = OD – OC CD = 5 – 3 CD = 2 Therefore, CD = 2 cm Hence, option A is the correct answer. |
|
| 53. |
AB and CD are two parallel chords of a circle which are on opposite sides of the centre such that `AB=10 cm`, `CD=24cm` and the distance between AB and CD is `17 cm`. Find the radius of the circle. |
| Answer» Correct Answer - `r=13 cm` | |
| 54. |
In the Figure, `O D`is perpendicular to the chord `A B`of a circle whose centre is `Odot`If `B C`is a diameter, show that `C A=2 OD` |
|
Answer» `OD_|_AB` D is mid point of AB BC is diameter O is the midpoint of BC BD=`1/2`AC AC=2*DB. |
|
| 55. |
In the adjoining figure ,AB and CD are two parallel chords of a circle with centre O , whose length are 16 cm and 12 cm respectively. Find the radius of the circle if the distance between them is 14 cm. |
| Answer» Correct Answer - 10 cm | |
| 56. |
In the adjoining figure, if chords AB and CD of the circle intesect each other at right angles, then find the value of `x+y`. |
| Answer» Correct Answer - `90^@` | |
| 57. |
In thefigure, `O D`isperpendicular to the chord `A B`of a circlewhose centre is `Odot`If `B C`is adiameter, show that `C A=2O Ddot` |
|
Answer» Since, `ODbotAB` (given) and the perpendicular drawn from the centre to a chord bisects the chord. `therefore` D is the mid -point of AB. Also, O is the mid -point of BC. Now, in `Delta ABC`, D and O are mid-points of AB and BC respectively. Therefore, `OD| |AC` and `OD = 1//2 AC`. or CA =2OD (mid-point theorem). |
|
| 58. |
In the adjoining figure. If two equal chords AB and CD of a circle intersect each other at E. then prove that chords AC and DB are equal. |
|
Answer» We have, `rArr` `oversetfrown(AB)congoversetfrown(CD)` (if two chords of a circle are equal then their corresponding arcs are congruent) Subtracting `oversetfrown(CB)` from both sides,we get `oversetfrown(AB)-oversetfrown(CB)congoversetfrown(CD)-oversetfrown(CB)` `rArr oversetfrown(AC)congoversetfrown(DB)` `rArr AC=DB` (if two arcs are congruent then their corresponding chords are equal). |
|
| 59. |
In fig., AB and CD are equal chords of a circle, with center O OM ⊥ AB and ON ⊥ CD then Prove that ∠OMN = ∠ONM |
|
Answer» Given : In circle C (O, r) chord AB = chord CD and OM ⊥ AB and ON ⊥ CD To prove : ∠OMN = ∠ONM Proof : We know that equal chord of a circle are equidistant from center. Thus OM = ON ΔOMN will be an isosceles triangle. In isosceles triangle angles opposite to equal sides are equal Thus ∠OMN = ∠ONM |
|
| 60. |
Chords equidistant to each other from center of circle are –(a) Double(b) Triple(c) Half(d) Equal |
|
Answer» Answer is (d) Equal |
|
| 61. |
Let C be the circle with centre (0, 0) and radius 3 units. The equation of the locus of the mid points of the chords of the circle C that subtend an angle of `(2pi)/3` at its center is (A)`x^2+y^2=3/2` (B) `x^2+y^2=1` (C) `x^2+y^2=27/4` (D) `x^2+y^2=9/4`A. `x^(2)+y^(2)=(3)/(2)`B. `x^(2)+y^(2)=1`C. `x^(2)+y^(2)=(27)/(4)`D. `x^(2)+y^(2)=(9)/(4)` |
|
Answer» Correct Answer - D |
|
| 62. |
Equal chords of a circle are equidistant from the centre. |
|
Answer» Given->AB=CD `OL_|_AB` and L is mid point of AB `AL=LB=1/2AB-(1)` `OM_|_CD` and M is midpoint of CD `CM=MD=1/2CD-(2)` `AB=CD` `1/2AB=1/2CD` `AL=CM` `In/_OMC and /_DLA` `OL=OA` `CM=AC` `/_OMC=/_OLA` `/_OMC cong /_DLA(RHS)`. |
|
| 63. |
In the figure, ray BC is tangent at point B and ray BA is secant. `/_ABC` intercepts are AXB if m(arc AXB ) `= 130^(@)` then find the measure of `/_ABC `. |
|
Answer» `/_ABC = (1)/(2) m (arc AXB) ` ….(Tangent second theorem ) `:. /_ABC = (1)/(2) xx 130= 65^(@)` |
|
| 64. |
Chords AB andCD of a circle with centre O are congruent. If m(arc AXB) `= 120^(@)` then what is the m(arc( CYD). |
|
Answer» chord `AB ~=` chord CD …..(given ) `m(arc AXB_ = m(arc CYD)` .....(Arcs corresponding to the congruent chords are congruent) m ( arc CYD ) = `120^(@)` |
|
| 65. |
In a circle with centre O, the measure of a minor arc is 110°. What is the measure of the major arc PYQ? |
|
Answer» Measure of major arc = 360° – measure of corresponding minor arc ∴ m (arc PYQ) = 360 – 110 ∴ m (arc PYQ) = 250° ∴ The measure of the major arc PYQ is 250°. |
|
| 66. |
If arc AXB and arc AYB are corresponding arcs and m(arc AXB) = 120° then m(arc AYB) = (A) 140° (B) 60° (C) 240° (D) 160° |
|
Answer» Correct answer is (C) 240° Hint: Measure of major arc = 360° – measure of corresponding minor arc ∴ m (arc AYB) = 360 – m (arc AXB) ∴ m (arc AYB) = 360 – 120 ∴ m (arc AYB) = 240° |
|
| 67. |
In an isosceles triangle `A B C`with `A B=A C ,`a circle passing through `B`and `C`intersects the sides `A Ba n dA C`at`Da n dE`respectively. Prove that `D E C Bdot` |
|
Answer» To prove`=/_ADE=/_ABC` Proof `/_ABC` is an isosceles `/_ACB=/_ABC-(1)` BCDE is a cyclic quadrilateral `/_ADE=/_ACB-(2)` From equation 1 and 2 `/_ABC=/_ADE`. |
|
| 68. |
The equation of the tangent to the circle `x^2+y^2=a^2,`which makes a triangle of area `a^2`with the coordinate axes, is`x+-y=asqrt(2)`(b) `x+-y=+-asqrt(2)``x+-y=2a`(d) `x+y=+-2a`A. `x+- y= +-a`B. `x+-y = +- a sqrt(2)`C. `x+-y=3a`D. `x+-y= +-2a` |
|
Answer» Correct Answer - 2 Let the tangent be of the form `(x)/(x_(1))+(y)/(y_(1))=1` and the area of triangle formed by it with the coordinate axes be `(1)/(2) x_(1)y_(2)=a^(2) ` (1) Using, the conditions of tangency , we get `(| - x_(1)y_(1)|)/(sqrt(x_(1)^(2)+y_(1)^(2)))=a` or `x_(1)^(2)+y_(1)^(2)=(x_(1)^(2)y_(1)^(2))/(a^(2))` (2) From (1) and (2) , we get the value of `x_(1)` and `y_(1)` , which gives the equation of tangent as `x+-y= +- a sqrt(2)`. |
|
| 69. |
lf a circle `C` passing through `(4,0)` touches the circle `x^2 + y^2 + 4x-6y-12 = 0` externally at a point `(1, -1),` then the radius of the circle `C` is :-A. 5B. `2sqrt5`C. `sqrt57`D. 4 |
|
Answer» Correct Answer - A Equation of tangent to the circle `x^(2)+y^(2)+4x-6y-12=0 " at " (1,-1)` is given by `"xx"_(1)+yy_(1)+2(x+x_(1))-3(y+y_(1))-12=0, " where " x_(1)= 1 and y_(1) = -1` `rArrx-y+2(x+1)-3(y-1)-12=0` `rArr 3x - 4y-7=0` This will also a tangent to the required circle. Now, equation of family of circles touching the line `3x-4y-7=0 " at point " (1,-1)` i s given bt `(x-1)^(2)+(y+1)^(2)+lambda (3x-4y-7)=0` So, the equation of required circle will be `(x-1)^(2)+(y+1)^(2)+lambda(3x-4y-7)=0, " for some " lambdainR" "...(i)` `therefore` The required circle passes through (4, 0) `therefore(4-1)^(2)+(0+1)^(2)+lambda(3xx4-4xx0-7)=0` `rArr 9+1+lambda(5)=0rArrlambda=-2` Subtituting `lambda = -2` in Eq. (i), we get `(x-1)^(2)+(y+1)^(2)-2(3x-4y-7)=0` `rArr x^(2) + y^(2)-8x+10y+16=0` On comparing it with `x^(2)+y^(2) +2gx+2fy+c=0`, we get `g=-4, f=5,c=16` `therefore "Radius"=sqrt(g^(2)+f^(2)-c)=sqrt(16+25+16)=5` |
|
| 70. |
Consider the family of circles `x^2+y^2-2x-2lambda-8=0`passing through two fixed points `Aa n dB`. Then the distance between the points `Aa n dB`is_____ |
|
Answer» Correct Answer - 6 `x^(2)+y^(2)-2x-2lambda y -8=0` or `(x^(2)+y^(2)-2x-8)-2 lambda y =0` which is of the form of `S + lambda L =0` All the circles pass through the point of intersection of the circle `x^(2)+y^(2)-2x-8=0` and `y=0` Solving, we gt `x^(2)-2x+8=0` or `(x-4)(x+2) =0` `:. A -= (4,0)` and `B -= (-2,0)` So, the distance between A and B is 6. |
|
| 71. |
A line meets the coordinate axes at A and B. A circle is circumscribed obout the triangle OAB. If the distance of the points A and B from the tangent at O, the origin, to the circle are m and n respectively, find the diameter of the circle.A. `(2m+n)/2`B. `(m+n)`C. `(mn)/(m+n)`D. `(m+2n)/2` |
|
Answer» Correct Answer - B |
|
| 72. |
The range of values of `alpha`for which the line `2y=gx+alpha`is a normal to the circle `x^2=y^2+2gx+2gy-2=0`for all values of `g`is`[1,oo)`(b) `[-1,oo)``(0,1)`(d) `(-oo,1]`A. `[1,oo)`B. `[-1,oo)`C. `(0,1)`D. `(- oo,1]` |
|
Answer» Correct Answer - 2 The line `2y=g x +alpha` should pass through `(-g, -g)` . So, `-2g = -g^(2)+alpha` or `alpha= g^(2)-2g = (g-1)^(2) - ge -1` |
|
| 73. |
Tangent to the curve `y=x^2+6` at a point `(1,7)` touches the circle `x^2+y^2+16x+12y+c=0 `at a point `Q`, then the coordinates of `Q` are (A) `(-6,-11)` (B) `(-9,-13)` (C) `(-10,-15)` (D) `(-6,-7)`A. 195B. 185C. 85D. 95 |
|
Answer» Correct Answer - D Key idea Equation of tangent to the curve `x^(2)= 4ay "at"(x_(1), y_(1)) is "xx"_(1)=4a((y+y_(1))/(2))` Tangent to the curve `x^(2)=y-6 "at" (1,7) " is "` `x = (y+7)/(2)-6` `rArr 2x-y+5=0" "(i)` Equation of circle is `x^(2)+y^(2)+16x + 12y+C=0 " Centre"(-8,-6)` `r=sqrt(8^(2)+6^(2)-c)=sqrt(100-c)` Since, line 2x - y + 5 = 0 also touches the circle. `therefore sqrt(100-c)=|(2(-8)-(-6)+5)/(sqrt(2^(2)+1^(2)))|` `rArr sqrt(100-c)=|(-16+6+5)/(sqrt5)|` `rArr sqrt(100-c)=|-sqrt5|` `rArr 100-c=5` `rArr c= 95` |
|
| 74. |
The tangent to the circle `x^2 + y^2 = 5` at (1, -2) also touches the circle `x^2 +y^2- 8x + 6y + 20=0`. Find the coordinates of the corresponding point of contact. |
|
Answer» Equation of tangent to `x^(2)+y^(2)=5` at (1,-2) is `x-2y-5=0`. Solving this with the second circle, we get `(2y+5)^(2)+y^(2)-8(2y+5)+6y+20=0` `implies 5y^(2)+10y+5=0` `implies (y+1)^(2)=0` `implies y=-1` `implies x =-2+5=3` Thus, point of contact on second circle is `(3,-1)`. |
|
| 75. |
Two circles of radii `aa n db`touching each other externally, are inscribed in the area bounded by `y=sqrt(1-x^2)`and the x-axis. If `b=1/2,`then `a`is equal to`1/4`(b) `1/8`(c) `1/2`(d) `1/(sqrt(2))`A. `(x-2)(x-8)+(y-4)(x-16)=0`B. `(1)/(8)`C. `(1)/(2)`D. `(1)/(sqrt2)` |
|
Answer» Correct Answer - `a=(1)/(4)` |
|
| 76. |
If the tangent at the point on the circle `x^2+y^2+6x +6y=2` meets the straight ine `5x -2y+6 =0` at a point Q on the y- axis then the length of PQ isA. 4B. `2sqrt(5)`C. 5D. `3 sqrt(5)` |
|
Answer» Correct Answer - 3 The line `5x-2y+6=0` is interested by a tangent at P to the circle `x^(2)+y^(2)+6x+6y-2=0` on the y-axis at `Q (0,3)`. In other words, the tangent passes through (0,3). Therefore, `PQ=` Length of tangent to circle from (0,3) `sqrt(0+9++18-2)` `=sqrt(25)=5` |
|
| 77. |
The locus of the point from which the lengths of the tangents to thecircles `x^2+y^2=4`and `2(x^2+y^2)-10 x+3y-2=0`are equal isa straight line inclined at `pi/4`with the line joining thecenters of the circlesa circle(c) an ellipsea straight line perpendicular to the line joining the centers of thecircles.A. a straight line inclined at `pi//4` with the line joining the centers of the circlesB. a circleC. an ellipseD. a straight line perpendicular to the line joining the centers of the circles |
|
Answer» Correct Answer - 4 The locus is the radical axis which is perpendicular to the line joining the centers of the circles. |
|
| 78. |
If the circles of same radius `a`and centers at (2, 3) and 5, 6) cut orthogonally, then find `adot`A. 1B. 2C. 3D. 4 |
|
Answer» Correct Answer - C |
|
| 79. |
Find the equation of the normals to the circle `x^2+y^2-8x-2y+12=0`at the point whose ordinate is `-1` |
|
Answer» We have circle `x^(2)+y^(2)-8x-2y+12=0` Centre of the circle is C(4,1). Putting `y= -1` in the equation of the circle, we get `x^(2)-8x+15=0` `implies(x-3)(x-5)=0` `implies x=5 ` or 3 Thus, the points on the circle are P(5,-1) and Q(3,-1). Equation of normal at P is `y+1=(-1-1)/(5-4)(x-5)` or `2x+y-9=0` Equation of normal at Q is `y+1=(-1-1)/(3-4)(x-5)` or `2x-y-7=0` |
|
| 80. |
Find the locus of the centers of the circles `x^2+y^2-2x-2b y+2=0`, where `a`and `b`are parameters, if the tangents from the origin to each of the circlesare orthogonal. |
|
Answer» The given circle is `x^(2)+y^(2)-2ax-2by+2=0` or `(x-a)^(2)+(y-b)^(2)=a^(2)+b^(2)-2` Its director circle is `(x-a)^(2)+(y-b)^(2)=2(a^(2)+b^(2)-2)` Given that tangents drawn from the origin to the circle are orthogonal. It implies that the director circle of the circle must pass through the origin, i.e., `a^(2)+b^(2)=2(a^(2)+b^(2)-2)` or `a^(2)+b^(2)=4` Thus, the locus of the center of the given circle is `x^(2)+y^(2)=4.` |
|
| 81. |
A pair of tangents are drawn from the origin to the circle `x^2 + y^2 + 20 (x + y) + 20 = 0`, The equation of pair of tangent is |
|
Answer» Correct Answer - `2x^(2)+2y^(2)+5xy=0` We have `S=x^(2)+y^(2)+20x+20y+20=0` Equation of pair of tangents from origin is given by `T^(2) S S_(1)`. or `(0.x+0.y+10(x+0)+10(y+0)+20)^(2)` `=(x^(2)+y^(2)+20x+20y+20)(20)` or `5(x+y+2)^(2)=(x^(2)+y^(2)+20x+20y+20)` or `2x^(2)+2y^(2)+5xy=0` |
|
| 82. |
The length of tangents from P(1,-1) and Q(3,3) to a circle are `sqrt2andsqrt6` respectively, then the length of tangent from R(-2,-7) to the same circle isA. `sqrt41`B. `sqrt51`C. `sqrt61`D. `sqrt71` |
|
Answer» Correct Answer - D |
|
| 83. |
Find the equation of pair of tangenst drawn to circle `x^(2)+y^(2)-2x+4y-4=0` from point P(-2,3). Also find the angle between tangest. |
|
Answer» We have circle `x^(2)+y^(2)-2x+4y-4=0` or `S=0`. Tangents are drawn from point P(-2,3) to the circle. Equation of pair of tangents is `T^(2)=S S_(1)` or `(-2x+3y-(x-2)+2(y+3)-4)^(2)` `=(x^(2)+y^(2)-2x+4y-4)((-2)^(2)+3^(2)-2(-2)+4(3)-4` `implies (-3x+5y+4)^(2)=25(x^(2)+y^(2)-2x+4y-4)` `implies 16x^(2)+30xy-26x+60y-116=0` `implies 8x^(2)+15xy-13x+30y-58=0` Here, a=8, b=0 and `h=15//2` If angle between tangents is `theta`, then `tan theta=(2 sqrt(h^(2)-ab))/(|a+b|)=(2sqrt((225)/(4)-0))/(|8+0|)=(15)/(8)` `:. theta =tan ^(-1).(15)/(8)` |
|
| 84. |
If the chord of contact of the tangents drawn from a point on thecircle `x^2+y^2+y^2=a^2`to the circle `x^2+y^2=b^2`touches the circle `x^2+y^2=c^2`, then prove that `a ,b`and `c`are in GP. |
|
Answer» Let (h,k) be the point `x^(2)+y^(2)=a^(2)`. Then, `h^(2)k^(2)=a^(2)` (1) The equation of the chord of contact of tangents drawn from (h,k) to `x^(2)+y^(2)=b^(2)` is `hx+ky=b^(2)` (2) This touches the circle `x^(2)+y^(2)=c^(2)`. Therefore, `|(-b^(2))/(sqrt(h^(2)+k^(2)))|` or `|(-b^(2))/(sqrt(a^(2)))|` [Using (1)] or `b^(2)=ac` Therefore, a,b, and c are in GP. |
|
| 85. |
If the chord of contact of tangents from a point `(x_1, y_1)` to the circle `x^2 + y^2 = a^2` touches the circle `(x-a)^2 + y^2 = a^2`, then the locus of `(x_1, y_1)` isA. a circleB. a parabolaC. an ellipseD. hyperbola |
|
Answer» Correct Answer - D |
|
| 86. |
Find the equation of a circle with Centre (a cos ∝, a sin ∝) and radius a |
|
Answer» The general form of the equation of a circle is: (x - h)2 + (y - k)2 = r2 Where, (h, k) is the centre of the circle. r is the radius of the circle. Substituting the centre and radius of the circle in he general form: (x - (a cos ∝))2 + (y - (a sin ∝))2 = a2 ⇒ (x - a cos ∝)2 + (y - a sin ∝)2 = a2 ⇒ x2 - 2xacos α + a2 cos2 α + y2 - 2yasin α + a2 sin2 α = a2 ⇒ x2 + y2 + a2 (cos2 α + sin2 α) - 2a(xcos α + ysin α) = a2 ⇒ x2 + y2 + a2 - 2a(xcos α + ysin α) = a2 …((cos2α + sin2α) = 1) ⇒ x2 + y2 - 2a(xcos α + ysin α) = 0 equation of a circle with Centre (a cos ∝, a sin ∝) and radius a is: x2 + y2 - 2a(xcos α + ysin α) = 0 |
|
| 87. |
A circle C of radius 1 is inscribed in an equilateral triangle PQR. The points of contact of C with the sides PQ, QR, RP are D, E, F, respectively. The line PQ is given by the equation `sqrt3 x+ y -6 = 0` and the point D is (3 sqrt3/2, 3/2). Further, it is given that the origin and the centre of C are on the same side of the line PQ. (1)The equation of circle C is (2)Points E and F are given by (3)Equation of the sides QR, RP areA. `y=(2)/(sqrt3)+x+1,y=-(2)/(sqrt3)x-1`B. `y=(1)/(sqrt3)x,y=0`C. `y=(sqrt3)/(2)x+1,y=-(sqrt3)/(2)x-1`D. `y=sqrt3x,y=0` |
|
Answer» Correct Answer - D |
|
| 88. |
A circle C of radius 1 is inscribed in an equilateral triangle PQR. The points of contact of C with the sides PQ, QR, RP are D, E, F, respectively. The line PQ is given by the equation `sqrt3 x+ y -6 = 0` and the point D is (3 sqrt3/2, 3/2). Further, it is given that the origin and the centre of C are on the same side of the line PQ. (1)The equation of circle C is (2)Points E and F are given by (3)Equation of the sides QR, RP areA. `((sqrt3)/(2),(3)/(2)),(sqrt3,0)`B. `((sqrt3)/(2),(1)/(2)),(sqrt3,0)`C. `((sqrt3)/(2),(1)/(2)),((sqrt3)/(2),(1)/(2))`D. `((3)/(2),(sqrt3)/(2)),((sqrt3)/(2),(1)/(2))` |
|
Answer» Correct Answer - A |
|
| 89. |
A circle C of radius 1 is inscribed in an equilateral triangle PQR. The points of contact of C with the sides PQ, QR, RP are D, E, F, respectively. The line PQ is given by the equation `sqrt3 x+ y -6 = 0` and the point D is (3 sqrt3/2, 3/2). Further, it is given that the origin and the centre of C are on the same side of the line PQ. (1)The equation of circle C is (2)Points E and F are given by (3)Equation of the sides QR, RP areA. `(x-2sqrt3)^(2)+(y-1)^(2)=1`B. `(x-2sqrt3)^(2)+(y+(1)/(2))^(2)=1`C. `(x-sqrt3)^(2)+(y+1)^(2)=1`D. `(x-sqrt3)^(2)+(y-1)^(2)=1` |
|
Answer» Correct Answer - D |
|
| 90. |
In the given figure, O is the centre of the circle. With reference to the figure fill in the blanks. |
|
Answer» 1. Seg OD is radius of the circle. 2. Seg AB is diameter of the circle. 3. Seg PQ is chord of the circle. 4. ∠DOB is the central angle. 5. Minor arc : arc AXD, arc BD, arc AP, arc PQ, arc BQ, etc. 6. Major arc : arc PAB, arc PDQ, arc PDB, arc ADQ, etc. 7. Semicircular arc : arc ADB, arc AQB. 8. m (arc DB) = m∠DOB 9. m (arc DAB) = 360° – m∠DOB |
|
| 91. |
If the circle `x^2+y^2=a^2`intersects the hyperbola `x y=c^2`at four points `P(x_1, y_1),Q(x_2, y_2),R(x_3, y_3),`and `S(x_4, y_4),`then`x_1+x_2+x_3+x_4=0``y_1+y_2+y_3+y_4=0``x_1x_2x_3x_4=C^4``y_1y_2y_3y_4=C^4`A. `x_(1)+x_(2)+x_(3)+x_(4)=0`B. `y_(1)+y_(2)+y_(3)+y_(4)=0`C. `x_(1)x_(2)x_(3)x_(4)=c^(4)`D. `y_(1)y_(2)y_(3)y_(4)=c^(4)` |
|
Answer» Correct Answer - 1,2,3,4 Putting `y=c^(2)//x` in `x^(2)+y^(2)=a^(2)`, we get `x^(2)+(c^(4))/(x^(2))=a^(2)` or `x^(4)-a^(2)x^(2)+c^(4)=0` As `x_(1),x_(2),x_(3)` and `x_(4)` are the roots of (i) ,we have `x_(1)+x_(2)+x_(3)+x_(4)=0` and `x_(1)x_(2)x_(3)x_(4)=c^(4)` Similarly, forming equation in , we get `y_(1)+y_(2)+y_(3)+y_(4)=0` and `y_(1)y_(2)y_(3)y_(4)=c^(4)` |
|
| 92. |
If the circles `x^2+y^2-9=0`and `x^2+y^2+2a x+2y+1=0`touch each other, then `alpha`is`-4/3`(b) 0(c) 1 (d)`4/3`A. `-4//3`B. 0C. 1D. `4//3` |
|
Answer» Correct Answer - 1,4 The equation of the radical axis is `2 alpha x +2y+10=0` i.e., `alpha x +y+5=0` (i) Putting the value of y from (i) in the circle `x^(2)+y^(2)=9`, we get `(1+alpha^(2))x^(2)+10 alpha x +16 =0` Radical axis is trangent . So, `D=0` or `36alpha^(2)-64=0` or `alpha = +- (4)/(3)` |
|
| 93. |
If the circle `x^2 + y^2 + ( 3 + sin beta) x + 2 cos alpha y = 0` and `x^2 + y^2 + 2 cos alpha x + 2 c y = 0` touch each other, then the maximum value of c is |
|
Answer» Correct Answer - 1 `x^(2)+y^(2)+(3+sin beta )x +(2 cos alpha) y =0` (1) `x^(2)+y^(2)+(2 cos alpha ) x +2cy =0` (2) Since both the circles are passing through the origin (0,0) , the equation of tangent to circles at (0,0) will be same . Tanget at (0,0) to circle (1) is, `(3 sin beta )x + (2 cos alpha ) y =0` (3) Tangent at (0,0) to circle (2) is `(2 cos alpha )x +2cy =0` Therefore, (1)/ and (2), we get `(3+sin beta)/(2 cos alpha )=(2c os alpha)/(2c)` or `c=(2cos^(2)alpha)/(3+sin beta)` `:. c_("max")=1` when `sin beta = -1 ` and `alpha =0` |
|
| 94. |
Difference in values of the radius of a circle whose center is at theorigin and which touches the circle `x^2+y^2-6x-8y+21=0`is_____________ |
|
Answer» Correct Answer - 4 Let r tbe the radius of the required circle. Now, if two circles touch each other, then Distance between their centers `= | r+- 2 | =5` (Given ) `:. R=3,7` |
|
| 95. |
Show that the circle x2 + y2 – 3x + 8y + 16 = 0 touches y – axis. |
|
Answer» Here f = 4, C = 16 & f2 = C ⇒ (4)2 = 16. ∴ Circle touches y-axis. |
|
| 96. |
If the circle `x^2+y^2-4x-8y-5=0`intersects the line `3x-4y=m`at two distinct points, then find the values of `mdot`A. `35 lt m lt 85`B. `-85 lt m lt - 35`C. `-35 lt m lt 15`D. `15 lt m lt 65` |
|
Answer» Correct Answer - C The circle is `x^(2)+y^(2)-4x-8y-5=0` Its centre is `(2,4)` and radius is `sqrt(4+16+5)=5` If the circle intersects the line `3x-4y=m` at two distinct points, then the length of the perpendicular from the centre is less than the radius, i.e., `(|6-16-m|)/(5) lt5` `implies |10+m| lt25` `implies 25 lt m +10lt25` `implies -35ltmlt15` |
|
| 97. |
The line `3x -4y = k` will cut the circle `x^(2) + y^(2) -4x -8y -5 = 0` at distinct points ifA. `-10ltlamdalt5`B. `9ltlamdalt20`C. `-35ltlamdalt15`D. `-16ltlamdalt30` |
|
Answer» Correct Answer - C |
|
| 98. |
If the circle `x^2+y^2-4x-8y-5=0`intersects the line `3x-4y=m`at two distinct points, then find the values of `mdot` |
|
Answer» Correct Answer - `-35 < m <15` We must have Radius of given circle `gt ` Perpendicular distance from the center of circle to the given line or `sqrt(4+16+5)gt (|3(2)-4(4)-m|)/(sqrt(9+16))` or `|m+10| lt 25` or `-35 lt m lt 15` |
|
| 99. |
If the line `3x-4y-lambda=0` touches the circle `x^2 + y^2-4x-8y- 5=0` at (a, b) then which of the following is not the possible value of `lambda+a + b`?A. `-22`B. `-20`C. 20D. 22 |
|
Answer» Correct Answer - C |
|
| 100. |
If the circle `x^2+y^2-4x-8y-5=0`intersects the line `3x-4y=m`at two distinct points, then find the values of `mdot`A. `-35ltmlt15`B. `15ltmlt65`C. `35ltmlt85`D. `-85ltmlt-35` |
|
Answer» Correct Answer - A |
|