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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.
| 251. |
`P Q`and `R S`are two parallel chords of a circle whose centre is `O`and radius is `10c mdot`If `P Q=16c m`and `R S=12c m ,`find the distance between `P Q`and `R S ,`if they lie:on the same side of the centre `Odot`on opposite side of the centre `Odot` |
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Answer» OP=OR=10cm PQ=14m,RS=12cm,PL=8cm,RM=6cm `In/_OLP` `OP^2=OL^2+PL^2` `100=OL^2+64` `OL^2=36` `OL=6` `In/_OMR` `OR^2=OM^2+RM^2` `100=OM^2+36` `OM^2=64` `OM=8` 1) On same side `LM=QM-OL` `LM=8-6=2cm` 2)On opposite side `LM=OL+QM` `LM=6+8=14cm`. |
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| 252. |
The lengths of the tangents from `P(1,-1)`and `Q(3,3)`to a circle are `sqrt(2)`and `sqrt(6)`, respectively. Then, find the length of the tangent from `R(-1,-5)`to the same circle. |
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Answer» Let the circle be `x^(2)+y^(2)+2gx+2fy+c=0`. Let the tangents from points P,Q,R touch the circle at A,B,C, respectively. According to question, `PA^(2)=2` or `1+1+2g-2f+c=2` or `2g-2f+c=0` (1) and `PB^(2)=5` or `9+9+6g+6f+c=6` or `6g+6f+c= -12` (2) From (1) and (2), `g= -1-(c)/(3),f=-1+(c)/(6)` Now, `RC^(2)=1+25-2g-10f+c` `=26+2+(2c)/(3)+10-(5c)/(3)+c` `=38` `:. RC =sqrt(38)` |
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| 253. |
Let `A-=(-1,0),B-=(3,0),`and `P Q`be any line passing through (4, 1) having slope `mdot`Find the range of `m`for which there exist two points on `P Q`at which `A B`subtends a right angle. |
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Answer» The equation of line PQ is `(y-1)=m(x-4)` or `y-mx+4m-1=0` For the required m, we have to make sure that the line PQ meets the circle, with diameter AB, at real and distinct points. The equation of the circle having AB as diameter is `x^(2)+y^(2)-2x-3=0` `:. (|0-m+4m-1|)/(sqrt(1+m^(2)))gt2` or `5m^(2)-6m-3lt0` or `m in ((3-2sqrt(6))/(5),(3+2sqrt(6))/(5))` |
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| 254. |
Locus of centre of a circle of radius `2,` which rolls on the outside of circle `x^2 +y^2 + 3x-6y-9=0` isA. `x^(2)+y^(2)+3x-6y-5=0`B. `x^(2)+y^(2)+3x-6y-31=0`C. `x^(2)+y^(2)+3x-6y-11=0`D. `x^(2)+y^(2)+3x-6y-36=0` |
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Answer» Correct Answer - B |
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| 255. |
Find the range of values of `m`for which the line `y=m x+2`cuts the circle `x^2+y^2=1`at distinct or coincident points. |
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Answer» Centre of the given circle is C(0,0) and radius is 1. Distance of centre of the circle from the given line is `CP=(|m(0)-0+2|)/(sqrt(1+m^(2)))=(2)/(sqrt(1+m^(2)))` If the line cuts the circle at two distinct or coincident points, then `CP lt1` `:. (2)/(sqrt(1+m^(2))) le1` `implies 1+m^(2)ge4` `implies m^(2)ge3` `implies m in(-oo,-sqrt(3)]cup[sqrt(3),oo)` |
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| 256. |
Find the greatest distance of the point `P(10 ,7)`from the circle `x^2+y^2-4x-2y-20=0`A. 5B. 10C. 15D. 20 |
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Answer» Correct Answer - C |
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| 257. |
The point `([p+1],[p])`is lying inside the circle `x^2+y^2-2x-15=0`. Then the set of all values of `p`is (where [.] represents the greatest integer function)`[-2,3)`(b) `(-2,3)``[-2,0)uu(0,3)`(d) `[0,3)`A. [-2,3]B. (-2,3)C. `[-2,3)cup(0,3)`D. [0,3) |
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Answer» Correct Answer - A |
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| 258. |
Find the greatest distance of the point `P(10 ,7)`from the circle `x^2+y^2-4x-2y-20=0` |
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Answer» Center of the given circle is C(2,1) and radius is `r=sqrt((-2)^(2)+(-1)^(2)-(-20))=5`. Now, `PC=sqrt((10-2)^(2)+(7-1)^(2))=10` Therefore, Greatest distance of point P from circle`=PC+r=10+5=15` Least distance of point P from circle `=|PC-r|=|10-5|=5` |
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| 259. |
The ends of a quadrant of a circle have the coordinates (1, 3) and (3,1). Then the center of such a circle is(2, 2) (b) (1, 1)(c) (4, 4)(d) (2, 6)A. (2,2)B. (1,1)C. (4,4)D. (2,6) |
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Answer» Correct Answer - 2 If `(alpha,beta)` is the center, then `(alpha-1)^(2)+(beta-3)^(2)=(alpha-3)^(2)+(beta-1)^(2)` (1) and `(beta-3)/(alpha-1).(beta-1)/(alpha-3)= -1` or `(alpha-1)(alpha-3)+(beta-1)(beta-3)=0` (2) From (1), `4 alpha -4beta=0` `:. alpha =beta` From (2), `2(alpha-1)(alpha-3)=0` `:. alpha=1,3` `:. (alpha,beta)-=(1,1),(3,3)` |
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| 260. |
Prove that the right bisector of a chord of a circle, bisects thecorresponding arc or the circle. |
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Answer» `/_OAC and /_OBC` `OA=OB` `QL=OL` `/_OLA=/_OLB=90^@` `/_OAC cong /_OBC(RHS)` `/_AOL=/_BOL` `AQ=BQ`. |
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| 261. |
P is a point on the circle `x^2+y^2=9` Q is a point on the line `7x+y+3=0`. The perpendicular bisector of PQ is `x-y+1=0`. Then the coordinates of P are:A. `(0,-3)`B. `(0,3)`C. `(72//25,21//25)`D. `(-72//25,21//25)` |
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Answer» Correct Answer - 3 Any point on the line `7x+y+3=0` is `Q(t ,-3-7t),t inR`. Now, `P(h,k)` is the image of point Q in the line `x-y+1=0`. Then, `(h-t)/(1)=(k-(-3-7t))/(-1)` `=-(2{t-(-3-7t)+1})/(1+1)` `=-8t-4` or `(h,k) -= (-7t-4,t+1)` This point lies on the circle `x^(2)+y^(2)=9`. Then, `(-7t-4)^(2)+(t+1)^(2)=9` or `50t^(2)+58t+8=0` or `25t^(2)+29t+4=0` or `(25t+4)(t+1)=0` or `t=(-4)/(25),t= -1` i.e., `(h,k) -= (-(72)/(25),(21)/(25))` or `(3,0)` |
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| 262. |
If the circumference of the circle `x^2 + y^2 + 8x + 8y - b = 0` is bisected by the circle `x^2 + y^2 - 2x + 4y + a = 0` then `a+b=` (A) 50 (B) 56 (C) `-56` (D) `-34`A. 50B. 56C. `-56`D. `-34` |
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Answer» Correct Answer - 3 The equation of radical axis ( i.e., common chord ) of the two circles is `10x+4ya-a-b=0` (1) The center of the first circle is `H(-4,-4)` Since the second circle bisects the circumference of the first circle, the center `H(-54,-4)` of the first circle must lie on the coomon chord (1). Therefore, ltbr. `-40-16-1-b=0` or `a+b= -56` |
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| 263. |
If the angle of intersection of the circle `x^2+y^2+x+y=0`and `x^2+y^2+x-y=0`is `theta`, then the equation of the line passing through (1, 2) and making anangle `theta`with the y-axis is`x=1`(b) `y=2``x+y=3`(d) `x-y=3`A. `x=1`B. `y=2`C. `x+y=3`D. `x-y=3` |
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Answer» Correct Answer - 2 Let A and B be the centers and `r_(1)` and `r_(2)` be the radii of the two circles. Then , `A-=(-(1)/(2),-(1)/(2)),B-=(-(1)/(2),(1)/(2)),` `r_(1)=(1)/(sqrt(2)),r_(2)=(1)/(sqrt(2))` `cos theta =(r_(1)^(2)+r_(2)^(2)-AB^(2))/(2r_(1)r_(2))` `=((1)/(2)+(1)/(2)-1)/(2xx(1)/(sqrt(2))xx(1)/(sqrt(2)))=0` or `theat =(pi)/(2)` Therefore, the required line is parallel to the x-axis and since it passes through (1,2), its equation will be y`=2` |
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| 264. |
From the point `P(sqrt(2),sqrt(6))`, tangents `P Aa n dP B`are drawn to the circle `x^2+y^2=4`Statement 1 :The area of quadrilateral `O A P B(O`being the origin) is 4.Statement 2 : The area of square is `a^2,`where `a`is the length of side.A. Statement I is true, Statement II is true, Statement II is a correct explanation for Statement IB. Statement I is true, Statement II is true, Statement II is not a correct explanation for Statement IC. Statement I is true, Statement II is falseD. Statement I is false, Statement II is true |
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Answer» Correct Answer - `:.` Both statements are true and statement II is correct explanation of statement. I |
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| 265. |
In figure, ∠DAB = 60°, ∠ABD = 50°, then ∠ACB equal(a) 60°(b) 50°(c) 70°(d) 80° |
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Answer» Answer is (c) 70° ∠DAB = 60° and ∠ADB = 50° In ΔADB ∠DAB + ∠ABD + ∠ADB = 180° ⇒ 60° + 50° + ∠ADB = 180° ⇒ ∠ADB = 180° – (60° – 50°) ⇒ ∠ADB = 70° ∠ACB and ∠ADB are angles in same segment. So will be equal ⇒ ∠ACB = ∠ADB = 70° |
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| 266. |
A point on the line `x=3` from which the tangents drawn to the circle `x^2+y^2=8` are at right angles isA. `(2,2sqrt(7))`B. `(2,2sqrt(5))`C. `(2,-2sqrt(7))`D. `(2,-2sqrt(5))` |
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Answer» Correct Answer - 1,3 The point from which the tangents drawn are at right angle lies on the director circle. Equation of director circle is `x^(2)+y^(2)=2 x 16 =32`. Putting `x=2` ,we get `y^(2)=28` or `y= +- 2sqrt(7)` |
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| 267. |
Prove that line joining the mid point of hypotenuse of right angled triangle to opposite vertex is half the hypotenuse. |
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Answer» Given : In right angled ΔABC ∠ABC = 90° and point M is mid point of hypotenuse AC To Prove : BM = AM = CM Construction : Taking AC as a diameter draw a circle which passes through B. Proof : Hypotenuse AC diameter circle, of right ΔABC, AC is M is mid point of AC so M will be center of circle Thus, AM = MC …(i) [M is mid point] BM = MC ….(ii) [Radius of circle] From equations (i) and (ii) BM = AM = MC = \(\frac { 1 }{ 2 }\)AC, |
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| 268. |
The two points `A` and `B` in a plane are such that for all points `P` lies on circle satisfied `PA/PB=k` , then `k` will not be equal to |
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Answer» Correct Answer - `k ne 1 ` Since, P lies on circle and A and B are points in plane such that, `(PA)/(PB)=k`, then the locus of P is perpendicular bisector of AB. Thus, the value of `k ne 1`. |
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| 269. |
The equation of the diameter of the circle `3(x^(2)+y^(2))-2x+6x-9-0` which is perpendicular to the line 2x+3y=12 isA. 3x-2y+3=0B. 3x-2y-3=0C. 3x-2y+1=0D. 3x-2y-1=0 |
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Answer» Correct Answer - B |
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| 270. |
Find the length of intercept, the circle `x^2+y^2+10 x-6y+9=0`makes on the x-axis. |
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Answer» Correct Answer - 8 Here, `g=5` and `c=9` `:. ` Length of intercept `=2 sqrt(g^(2)-c)=2 sqrt(25-9)=8` |
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| 271. |
The line `x+3y=0`is a diameter of the circle `x^2+y^2-6x+2y=0` |
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Answer» Correct Answer - 1 Since, centre of circle is (3, -1) which lies on x + 3y = 0 `rArr x + 3y = 0 " is diameter of " x^(2) + y^(2)-6x + 2y = 0` Hence, given statement is true. |
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| 272. |
If the line `x+2b y+7=0`is a diameter of the circle `x^2+y^2-6x+2y=0`, then find the value of `b` |
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Answer» Correct Answer - `b=5` Here the center of the circle, (3,-1) , must lie on the line `x+2by+7=0`. Therefore, `3-2b+7=0` or `b=5` |
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| 273. |
If the line `x+2b y+7=0`is a diameter of the circle `x^2+y^2-6x+2y=0`, then find the value of `b`A. 1B. 3C. 5D. 7 |
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Answer» Correct Answer - C |
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| 274. |
Find the equation of a circle concentric with the circle ` 2x ^(2) + 2y ^(2) - 6x + 8y + 1 = 0 ` and of double its area. |
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Answer» The given equation of the circle is ` 2x ^(2) + 2 y ^(2) - 6x + 8y + 1 = 0 ` ` rArr x ^(2) + y ^(2) - 3x + 4y + (1)/(2) = 0 rArr x ^(2) + y ^(2) + 2gx + 2f y + c = 0`, where ` g = (-3)/(2), f = 2 and c = (1)/(2)` . Centre of this circle `= (-g, - f ) = ((3)/(2), - 2 )` Radius of this circle `= sqrt (g ^(2) + f ^(2) - c )` ` " " = sqrt ((9)/(4) + 4 - (1)/(2)) = (sqrt ( 23 ))/( 2)` ` therefore ` the centre of the required circle ` = ((3)/(2), - 2)`. Let ` r_1 ` be the radius of the required circle. Then, `pi r _1 ^(2) = 2 { pi xx ((sqrt ( 23))/(2)) ^(2) } rArr r_1 ^(2) = ( 23)/(2)` Hence , the equation of the required circle is ` (x - (3)/ ( 2) ) ^(2) + (y + 2 ) ^(2) = ( 23)/(2)` ` rArr x ^(2) + y ^(2) - 3x + 4y - ( 21)/(4) = 0 ` ` rArr 4x ^(2) + 4y ^(2) - 12 x + 16 y - 21 = 0 ` |
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| 275. |
If one end of a diameter of the circle `2x^(2)+2y^(2)-4x-8y+2=0` is (-1,2), then the other end of the diameter isA. (2,1)B. (3,2)C. (4,3)D. (5,4) |
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Answer» Correct Answer - B |
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| 276. |
If one end of the a diameter of the circle `2x^2+2y^2-4x-8y+2=0`is (3, 2), then find the other end of the diameter. |
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Answer» Correct Answer - (-1,2) The center of the given circle is (1,2). Let (alpha,beta)` be the other end. Then, `=2 sqrt(g^(2)-c)=2 sqrt(25-9)=8` |
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| 277. |
Number of integral values of `lambda` for which `x^2 + y^2 + 7x + (1-lambda)y + 5 = 0` represents the equation of a circle whose radius cannot exceed 5 is |
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Answer» Correct Answer - 10 values Radius `le5`. `sqrt(((lambda)/(2))^(2)+((1-lambda)/(2))^(2)-5)le5` `implies 2lambda^(2)-2lambda-119le0` `implies (1-sqrt(239))/(2)lelambdale(1+sqrt(239))/(2)` `implies -7.2 lelambdale8.2 ` ( approximately) `implies lambda=-7,-6,-5.....7,8` (1) Also, we must have `((lambda)/(2))^(2)+((1-lambda)/(2))^(2)-5ge0` `implies 2lambda^(2)-2lambda-19ge0` `implies lambdale(1-sqrt(39))/(2)` or `lambdale(1+sqrt(39))/(2)` (2) From (1) and (2) , `lambda = -7,-6,-5,-4,-3,4,5,6,7,8` |
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| 278. |
Find the radius of the circle `(x-5)(x-1)+(y-7)(y-4)=0`. |
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Answer» Correct Answer - `5//2` The extremities of diameter are (5,7) and (1,4). The radius is half of the distance between them, i.e., `(1)/(2) sqrt((5-1)^(2)+(7-4)^(2))=(5)/(2)` |
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| 279. |
Find the locus of center of circle of radius 2 units, if intercept cuton the x-axis is twice of intercept cut on the y-axis by the circle. |
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Answer» Correct Answer - `4x^(2)-y^(2)=12` Let the circle be `x^(2)+y^(2)+2gx+2fy+c=0` Therefore , the intercept on the x-axis is `2 sqrt(g^(2)-c)` Also, the intercept on the y-axis is `2 sqrt(f^(2)-c)` According to the question, `sqrt(g^(2)-c)=2sqrt(f^(2)-c)` or `g^(2)-4f^(2)= -3c` (1) Given, radius `=2` `:. g^(2)+f^(2)-c=4` (2) Eliminating c from (1) and (2), we get `4g^(2)-f^(2)=12` Therfore, the locus is `4x^(2)-y^(2)=12` |
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| 280. |
For each natural number k, let `C_k` denotes the circle radius k centimeters in the counter-clockwise direction.After completing its motion on `C_k`, the particle moves to `C_[k+1]` in the radial direction. The motion of the particle continues in this manner. The particle starts at (1,0).If the particle crosses the the positive direction of the x-axis for first time on the circle `C_n`,then n equal to |
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Answer» Correct Answer - n = 7 It is given that , `C_(1) " has centre " (0, 0)` and radius 1. Similarly, `C_(2)` has centre (0, 0) and radius 2 and `C_(k)` ahs centre (0, 0) and radius k. Now, particle starts it motion from (1, 0) and moves 1 radian on first circle then particle shifts from `C_(1)` to` C_(2)`. After that, particle moves 1 radian on `C_(2)` and then paricle shifts from `C_(2) " to " C_(3)`. Similarly, particle move on n circles. Now, `n ge 2pi` because particle crosses the X-axis for the first time on `C_(n)`, then n is least positive integer. therefore, n = 7. |
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| 281. |
The intercept on the line `y=x` by the circle `x^2+y^2-2x=0` is AB. Equation of the circle with AB as a diameter is (A) `(x-1/2)^3+(y-1/2)^2=1/2` (B) `(x-1/2)^2+(y-1/2)^2=1/4` (C) `(x+1/2)^2+(y+1/2)^2=1/2` (D) `(x+1/2)^2+(y+1/2)^2=1/4` |
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Answer» Correct Answer - `x^(2)+y^(2)-x-y=0` Equation of any circle passing through the point of lt brgt intersection of `x^(2)+y^(2)-2x=0 and y=x`is `(x^(2)+y^(2)-2x)+lambda(y-x)=0` `rArr x^(2)+y^(2)-(2+lambda)x+lambday=0` Its centre is `((2+lambda)/(2),(-lambda)/(2))`. `rArr` For AB to be the diameter of the required circle the centre must be on AB, i.e. `2+lambda=-lambda` `rArr lambda = -1 " "[therefore y = x]` therefore, equation of the required circle is `x^(2)+y^(2)=(2-1)x-1*y=0` `rArr x^(2)+y^(2)-x-y=0` |
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| 282. |
The locus of the centre of the circle`(xcos alpha + y sin alpha - a)^2 + (x sin alpha - y cos alpha - b)^2= k^2` if `alpha` varies, isA. `x^(2)-y^(2)=a^(2)+b^(2)`B. `x^(2)-y^(2)=a^(2)b^(2)`C. `x^(2)+y^(2)=a^(2)+b^(2)`D. `x^(2)+y^(2)=a^(2)b^(2)` |
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Answer» Correct Answer - 3 `(x cos alpha+y sing alpha -a)^(2)-(x sin alpha-y cos alpha-b)^(2)=k^(2)` or `x^(2)+y^(2)+(-2a cos alpha-2b sin alpha)x+(-2a cos alpha+2b sin alpha)y+a^(2)+b^(2)-k^(2)=0` Centre of the circle is `(a cos alpha +b sin alpha,a cos alpha-b sin alpha) -= (h,k).` `h-=a cos alpha +b sin alpha` `k -= a cos alpha - b sin alpha` Squaring and adding, we get `h^(2)+k^(2)=a^(2)+b^(2)` or `x^(2)+y^(2)=a^(2)+b^(2)`, which is required locus. |
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| 283. |
If it is possible to draw a triangle which circumscribes the circle `(x-(a-2b))^2+(y-(a+b))^2=1` and is inscribed by `x^2+y^2-2x-4y+1=0` thenA. `beta = -(1)/(3)`B. `beta =(2)/(3)`C. `alpha=(5)/(3)`D. `alpha = - (5)/(2)` |
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Answer» Correct Answer - 3 For such triangle, first circle is the incircle and other one is the circumcircle. Also, the radius of first circle is half of the radius of the second circle. So, triangle is equilateral Thus, incentre and circumcentre coincide. `implies alpha-2beta=1` and `alpha+beta=2` `implies (alpha,beta)-=(5)/(3),(1)/(3)` |
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| 284. |
`f(x , y)=x^2+y^2+2a x+2b y+c=0`represents a circle. If `f(x ,0)=0`has equal roots, each being `2,`and `f(0,y)=0`has 2 and 3 as its roots, then the center of the circle is`(2,5/2)`(b) Data are not sufficient`(-2,-5/2)`(d) Data are inconsistentA. `(2,5//2)`B. Data are not sufficientC. `(-2,-5//2)`D. Data are inconsistent. |
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Answer» Correct Answer - 3 `x^(2)+2ax+c=(x-2)^(2)` or `-2a=4,c=4` or `a= -2,c=4` `y^(2)+2by+c=(y-2)(y-3)` or `-2b=5,c=6` or `b=- (5)/(2),c=6` Clearly, the data are not consistent. |
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| 285. |
In the adjoining figure, BP is the diameter of the circle. If `angle ABP=60^(@)`, then find `angle AQB`. |
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Answer» BP is the diameter of the circle. `thereforeangle BAP=90^@` In `Delta ABP`, `angle APB=180^@-(angle BAP+angleABP)` `=180^@-(90^@+60^@)` `30^@` (angles of same segment ) Now, `angle AQB=angle APB` `rArr angle AQB=30^@` |
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| 286. |
In the given figure, find `angle ADB`, if `angle DCB=100^@` and ` angle DBA =70^@`. |
| Answer» Correct Answer - `30^@` | |
| 287. |
In the adjoining figure, O is the centre of the centre of the circle. If diameter AC=26cm and chord AB=10cm, then find the distances of the chord AB from the centre of the circle. |
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Answer» Radius of circle `=("diameter")/(2)` `rArr AO=(26)/(2)=13cm` `AM=(AB)/(2)=(10)/(2)=5cm, (because "perpendicular drawn from centre to the chrod bisects the chord" )` Now, in `Delta AOM`, `AM^2+OM^2=AO^2` `rArrOM^2=AO^2-AM^2=13^2-5^2=169-25=144` `rArrOM=sqrt(144)=12 cm` Therefore, the distance of chord from the centre =12 cm |
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| 288. |
The circle passing through the point (-1,0) and touching the y-axis at (0,2) also passes through the point:A. `( -3//2,0)`B. `(- 5//2,2)`C. `(-3//2,5//2)`D. `(-4,0)` |
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Answer» Correct Answer - D Equation of circle touching the y-axis at (0,2) is `(x-0)^(2)+(y-2)^(2)+lambdax =0` It passes through `(-1,0)` . Therefore, `1+4-lambda =0` or `lambda =5` `:. `Equation of circle is `x^(2)+y^(2)+5x-4y+4=0` Putting `y=0` , we get ` x= -1,-4` Therefore, the circle passes through `(-4,0)` |
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| 289. |
Find the centre and radius of each of the following circles : (x – 3)2 + (y – 1)2 = 9 |
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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. Comparing the given equation of circle with general form we get: h = 3 , k = 1, r2 = 9 ⇒ centre = (3, 1) and radius = 3 units. Ans: centre = (3, 1) and radius = 3 units |
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| 290. |
Find the equation of the circle passing through the points :(i) (5, 7), (8, 1) and (1, 3)(ii) (1, 2), (3, – 4) and (5, – 6)(iii) (5, -8), (-2, 9) and (2, 1)(iv) (0, 0), (-2, 1) and (-3, 2) |
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Answer» (i) (5, 7), (8, 1) and (1, 3) By using the standard form of the equation of the circle: x2 + y2 + 2ax + 2by + c = 0 ….. (1) Firstly let us find the values of a, b and c Substitute the given point (5, 7) in equation (1), we get 52 + 72 + 2a (5) + 2b (7) + c = 0 25 + 49 + 10a + 14b + c = 0 10a + 14b + c + 74 = 0….. (2) By substituting the given point (8, 1) in equation (1), we get 82 + 12 + 2a (8) + 2b (1) + c = 0 64 + 1 + 16a + 2b + c = 0 16a + 2b + c + 65 = 0….. (3) Substituting the point (1, 3) in equation (1), we get 12 + 32 + 2a (1) + 2b (3) + c = 0 1 + 9 + 2a + 6b + c = 0 2a + 6b + c + 10 = 0….. (4) Now by simplifying the equations (2), (3), (4) we get the values a = -29/6, b = -19/6, c = 56/3 Substituting the values of a, b, c in equation (1), we get x2 + y2 + 2 (-29/6)x + 2 (-19/6) + 56/3 = 0 x2 + y2 – 29x/3 – 19y/3 + 56/3 = 0 3x2 + 3y2 – 29x – 19y + 56 = 0 ∴ The equation of the circle is 3x2 + 3y2 – 29x – 19y + 56 = 0 (ii) (1, 2), (3, – 4) and (5, – 6) By using the standard form of the equation of the circle: x2 + y2 + 2ax + 2by + c = 0 ….. (1) Substitute the points (1, 2) in equation (1), we get 12 + 22 + 2a (1) + 2b (2) + c = 0 1 + 4 + 2a + 4b + c = 0 2a + 4b + c + 5 = 0….. (2) Substitute the points (3, -4) in equation (1), we get 32 + (- 4)2 + 2a (3) + 2b (- 4) + c = 0 9 + 16 + 6a – 8b + c = 0 6a – 8b + c + 25 = 0….. (3) Substitute the points (5, -6) in equation (1), we get 52 + (- 6)2 + 2a (5) + 2b (- 6) + c = 0 25 + 36 + 10a – 12b + c = 0 10a – 12b + c + 61 = 0….. (4) Now by simplifying the equations (2), (3), (4) we get a = – 11, b = – 2, c = 25 Substitute the values of a, b and c in equation (1), we get x2 + y2 + 2(- 11)x + 2(- 2) + 25 = 0 x2 + y2 – 22x – 4y + 25 = 0 ∴ The equation of the circle is x2 + y2 – 22x – 4y + 25 = 0 (iii) (5, -8), (-2, 9) and (2, 1) By using the standard form of the equation of the circle: x2 + y2 + 2ax + 2by + c = 0 ….. (1) Substitute the point (5, -8) in equation (1), we get 52 + (- 8)2 + 2a(5) + 2b(- 8) + c = 0 25 + 64 + 10a – 16b + c = 0 10a – 16b + c + 89 = 0….. (2) Substitute the points (-2, 9) in equation (1), we get (- 2)2 + 92 + 2a(- 2) + 2b(9) + c = 0 4 + 81 – 4a + 18b + c = 0 -4a + 18b + c + 85 = 0….. (3) Substitute the points (2, 1) in equation (1), we get 22 + 12 + 2a(2) + 2b(1) + c = 0 4 + 1 + 4a + 2b + c = 0 4a + 2b + c + 5 = 0….. (4) By simplifying equations (2), (3), (4) we get a = 58, b = 24, c = – 285. Now, by substituting the values of a, b, c in equation (1), we get x2 + y2 + 2(58)x + 2(24) – 285 = 0 x2 + y2 + 116x + 48y – 285 = 0 ∴ The equation of the circle is x2 + y2 + 116x + 48y – 285 = 0 (iv) (0, 0), (-2, 1) and (-3, 2) By using the standard form of the equation of the circle: x2 + y2 + 2ax + 2by + c = 0 ….. (1) Substitute the points (0, 0) in equation (1), we get 02 + 02 + 2a(0) + 2b(0) + c = 0 0 + 0 + 0a + 0b + c = 0 c = 0….. (2) Substitute the points (-2, 1) in equation (1), we get (- 2)2 + 12 + 2a(- 2) + 2b(1) + c = 0 4 + 1 – 4a + 2b + c = 0 -4a + 2b + c + 5 = 0….. (3) Substitute the points (-3, 2) in equation (1), we get (- 3)2 + 22 + 2a(- 3) + 2b(2) + c = 0 9 + 4 – 6a + 4b + c = 0 -6a + 4b + c + 13 = 0….. (4) By simplifying the equations (2), (3), (4) we get a = -3/2, b = -11/2, c = 0 Now, by substituting the values of a, b, c in equation (1), we get x2 + y2 + 2(-3/2)x + 2(-11/2)y + 0 = 0 x2 + y2 – 3x – 11y = 0 ∴ The equation of the circle is x2 + y2 – 3x – 11y = 0 |
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| 291. |
Find the centre and radius of the following circles: x2 + y2 = 25 |
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Answer» Given equation of the circle is x2 + y2 = 25 ⇒ x2 + y2 = (5)2 Comparing this equation with x2 + y2 = r2 , we get r = 5 Centre of the circle is (0, 0) and radius of the circle is 5. |
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| 292. |
Find the centre and radius of each of the following circles : \(\Big(x-\frac{1}{2}\Big)^2+\Big(y+\frac{1}{3}\Big)^2=\frac{1}{16}\) |
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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. Comparing the given equation of circle with general form we get: h = 1/2 , k = - 1/3, r2 = 1/16 ⇒ centre = (1/2, - 1/3) and radius = 1/4 units. Ans: centre = (1/2, - 1/3) and radius = 1/4 units. |
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| 293. |
Find the equation of the circle which passes through (3, – 2), (- 2, 0) and has its centre on the line 2x – y = 3. |
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Answer» Given: The line 2x – y = 3 … (1) The points (3, -2), (-2, 0) By using the standard form of the equation of the circle: x2 + y2 + 2ax + 2by + c = 0 ….. (2) Let us substitute the centre (-a, -b) in equation (1) we get, 2(- a) – (- b) = 3 -2a + b = 3 2a – b + 3 = 0…… (3) Now Substitute the given points (3, -2) in equation (2), we get 32 + (- 2)2 + 2a(3) + 2b(- 2) + c = 0 9 + 4 + 6a – 4b + c = 0 6a – 4b + c + 13 = 0….. (4) Substitute the points (-2, 0) in equation (2), we get (- 2)2 + 02 + 2a(- 2) + 2b(0) + c = 0 4 + 0 – 4a + c = 0 4a – c – 4 = 0….. (5) By simplifying the equations (3), (4) and (5) we get, a = 3/2, b = 6, c = 2 Again by substituting the values of a, b, c in (2), we get x2 + y2 + 2 (3/2)x + 2 (6)y + 2 = 0 ∴ The equation of the circle is x2 + y2 + 3x + 12y + 2 = 0. |
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| 294. |
Find the centre and radius of the following circles: (x – 5)2 + (y – 3)2 = 20 |
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Answer» Given equation of the circle is (x – 5)2 + (y – 3)2 = 20 ⇒ (x – 5)2 + (y – 3)2 = (√20)2 Comparing this equation with (x – h)2 + (y – k)2 = r2 , we get h = 5, k = 3 and r = √20 = 2√5 Centre of the circle = (h, k) = (5, 3) and radius of the circle = 2√5. |
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| 295. |
Find the centre and radius of each of the following circles : (x + 5)2 + (y – 3)2 = 20 |
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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. Comparing the given equation of circle with general form we get: h = - 5 , k = 3, r2 = 20 ⇒ centre = ( - 5, 3) and radius = √20 = 2√5 units. Ans: centre = ( - 5, 3) and radius = 2√5 units. |
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| 296. |
Show that the quadrilateral formed by the straight lines `x-y = 0,3x+2y=5, x-y=10 and 2x + 3y = 0` is cyclic and hence find the equation of the circle. |
| Answer» Correct Answer - ` x ^(2) + y ^(2) - 6x + 4y = 0 ` | |
| 297. |
Find the centre and radius of each of the following circles : x2 + (y – 1)2 = 2 |
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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. Comparing the given equation of circle with general form we get: h = 0 , k = 1, r2 = 2 ⇒ centre = (0, 1) and radius = √2units. Ans: centre = (0, 1) and radius = √2units. |
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| 298. |
Find the equation of the circle which passes through the points (3, 7), (5, 5) and has its centre on line x – 4y = 1. |
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Answer» Given: The points (3, 7), (5, 5) The line x – 4y = 1…. (1) By using the standard form of the equation of the circle: x2 + y2 + 2ax + 2by + c = 0 ….. (2) Let us substitute the centre (-a, -b) in equation (1) we get, (- a) – 4(- b) = 1 -a + 4b = 1 a – 4b + 1 = 0…… (3) Substitute the points (3, 7) in equation (2), we get 32 + 72 + 2a(3) + 2b(7) + c = 0 9 + 49 + 6a + 14b + c = 0 6a + 14b + c + 58 = 0….. (4) Substitute the points (5, 5) in equation (2), we get 52 + 52 + 2a(5) + 2b(5) + c = 0 25 + 25 + 10a + 10b + c = 0 10a + 10b + c + 50 = 0….. (5) By simplifying equations (3), (4) and (5) we get, a = 3, b = 1, c = – 90 Now, by substituting the values of a, b, c in equation (2), we get x2 + y2 + 2(3)x + 2(1)y – 90 = 0 x2 + y2 + 6x + 2y – 90 = 0 ∴ The equation of the circle is x2 + y2 + 6x + 2y – 90 = 0. |
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| 299. |
Find the equation of the circle with centre at (2, 1) and passing through (0, -1). |
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Answer» Given Centre = C (2, 1) & Let P = (0, -1) Here r = CP = \(\sqrt{(0 - 2)^2 + (-1 - 1)^2} = \sqrt{4 + 4} \) = \(\sqrt 8 \) units Equation of the circle with centre (2, 1) & r =(\(\sqrt 8 \))2 is (x – 2)2 + (y – 1)2 = \(\sqrt 8 \) x2 + 4x + y2 + 1 – 2y = 8 x2 + y2 – 4x – 2y – 3 = 0. |
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| 300. |
Find the equation of the circle which passes through the points (1, 3) and (2, - 1), and has its centre on the line 2x + y – 4 = 0. |
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Answer» The equation of a circle: x2 + y2 + 2gx + 2fy + c = 0…(i) Putting (1, 3) & (2, - 1) in (i) 2g + 6f + c = - 10..(ii) 4g - 2f + c = - 5..(iii) Since the centre lies on the given straight line, ( - g, - f) must satisfy the equation as - 2g –f - 4 = 0…(iv) Solving, f = - 1, g = - 1.5, c = - 1 The equation is x2 + y2 - 3x - 2y - 1 = 0 |
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