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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.
| 401. |
In the adjoining figure, BD is the diameter of the circle which bisects the chord AC at point E. If `AC=8cm`, `BE=2cm`, then find the radius of the circle. |
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Answer» Let radius `OA=OB =r` `OE=OB-EB` `=(r-2)cm` `AE=(AC)/(2)=(8)/(2)=4cm` Now, `angle OEA=90^@` (`because` The line segment joining the centre to mid-point of a chord is perpendicular to the chord) `therefore` In `DeltaOAE`, `OA^2=OE^2+AE^2` (by Pythagoras theorem) `rArrr^2=(r-2)^2+4^2` `rArrr^2=r^2-4r+4+16` `rArr4r=20 cm` `rArr=5cm` `therefore` Radius of circle =5 cm . |
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| 402. |
If diameter of a circle bisects two chords, then chords will be –(a) Parallel(b) Perpendicular(e) Intersecting(d) None of there |
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Answer» Answer is (a) Parallel |
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| 403. |
The exhaustive range of value of a such that the angle between the pair of tangents drawn from `(a,a)` to the circle `x^2 +y^2 -2x -2y -6 =0` lies in the range `(pi/3,pi)` isA. (-1,3)B. `(-5,-3)uu(3,5)`C. `(-3,5)`D. `(-3,-1)uu(3,5)` |
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Answer» Correct Answer - D |
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| 404. |
One of the diameter of a circle circumscribing the rectangle ABCD is `4y = x + 7`, If A and B are the points `(-3, 4)` and `(5, 4)` respectively, then the area of rectangle isA. 16 sq unitsB. 24 sq unitsC. 32 sq unitsD. None of these |
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Answer» Correct Answer - C |
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| 405. |
The possible value of `lamda(lamdagt0)` such that the angle between the pair of tangents from point `(lamda,0)` to the circle `x^(2)+y^(2)=4` lies in interval `((pi)/(2),(2pi)/(3))` isA. `((4)/(sqrt3),2sqrt2)`B. `(0,sqrt2)`C. (1,2)D. `(-(4)/(sqrt3),(4)/(sqrt3))` |
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Answer» Correct Answer - A |
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| 406. |
Let `phi(x,y)=0` be the equation of a circle. If `phi(0,lamda)=0` has equal roots `lamda=2,2fandphi(lamda,0)=0` has roots `lamda=(4)/(5),5` then the centre of the circle isA. `(2,(29)/(10))`B. `((29)/(10),2)`C. `(-2,(29)/(10))`D. None of these |
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Answer» Correct Answer - B |
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| 407. |
If P(2,8) is an interior point of a circle `x^(2)+y^(2)-2x+4y-lamda=0` which neither touches nor intersects the axes, then set for `lamda` isA. `(-oo,-1)`B. `(-oo,-4)`C. `(96,oo)`D. `phi` |
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Answer» Correct Answer - D |
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| 408. |
Let a circle be given by `2x(x-1)+y(2y-b)=0,(a!=0,b!=0)`. Find the condition on `aa n db`if two chords each bisected by the x-axis, can be drawn to the circlefrom `(a , b/2)` |
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Answer» The given circle can be rewritten as `x^(2)+y^(2)-ax-(by)/(2)=0" "...(i)` Let one of the chord through (a, b/2) be bisected at (h, 0). Then, the equation of the chord having (h, 0) as mid-point is 0 `T=S_(1)` `rArr h*x+0*y-(a)/(2)(x+h)-(b)/(h)(y+0)=h^(2)+0-ah-0` `rArr (h-(a)/(2))x-(by)/(4)-(a)/(2)h=h^(2)-ah" "...(ii)` It passes through `(a, b//2)`,then `(h-(a)/(2))a-(b)/(4)*(b)/(2)-(a)/(2)=h^(2)-ah` According to the given condition, Eq. (iii) must have tow distinct real roots. This is possible, if the discriminant of Eq. (iii) is greater than 0. i.e. `(9)/(4)a^(2)-4((a^(2))/(2)+b^(2)/8)gt0rArr (a^(2))/(4)-(b^(2))/(2)gt0` `rArr a^(2)gt2b^(2)` |
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| 409. |
The values of `lamda` for which the circle `x^(2)+y^(2)+6x+5+lamda(x^(2)+y^(2)-8x+7)=0` dwindles into a point areA. `1pm(sqrt2)/(3)`B. `2pm(2sqrt2)/(3)`C. `2pm(4sqrt2)/(3)`D. `1pm(4sqrt2)/(3)` |
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Answer» Correct Answer - `lamda=2pm(4sqrt2)/(3)` |
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| 410. |
If two distinct chords, drawn from the point (p, q) on the circle `x^2+y^2=p x+q y`(where `p q!=q)`are bisected by the x-axis, then`p^2=q^2`(b) `p^2=8q^2``p^28q^2`A. `|p|=|q|`B. `p^(2)=8q^(2)`C. `p^(2)lt8q^(2)`D. `p^(2)gt8q^(2)` |
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Answer» Correct Answer - `p^(2)gt8q^(2)` |
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| 411. |
Given `:` `square ` ABCD is cyclic. `/_DCE ` is an exterior angle of `square ` ABCD. To Prove `: /_DCE = /_BAD ` Complete the proof by filling the boxes. |
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Answer» `/_DCE + /_BCD = square` …(Linear pair of angles ) ….(1) `square ` ABCD is cyclic `/_BAD + square = 180^(@)` …(Theorem of `square ` ) ….2) `:.` from (1) and (2), we get, `/_DCE + square = square + /_BCD ` `:. /_DCE =square ` Activity `:` ` /_ DCE + /_BCD = 180^(@)` ....(Linear pair of angles ) ...(1) `square ` ABCD is cyclic `/_ BAD + /_BCD = 180^(@)` ...(Theorem of Cyclic quadrilateral ) ...(2) `:.` from (1) and (2), we get `/_ DCE + /_BCD = /_BAD + /_BCD ` `:. /_DCE = /_BAD ` |
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| 412. |
The sum of the angles in the four segments exterior to a cyclicquadrilateral is equal to `6`right angles. |
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Answer» APBS iis a cyclic quadrilateral `/_1+/_P=180^@-(1)` BQCS is a cyclic quadrilateral `/_2+/_Q=180^@-(2)` CRQS is a cyclic quadrilateral `/_3+/_R=180^@-(3)` `/_P+/_Q+/_1+/_2+/_R+/_3=3*180^@` `/_P+/_Q+/_R+/_1+/_2+/_3=3*180^@` `/_P+/_Q+/_R+/_S=1*180^@` `/_P+/_Q+/_R+/_S=6`right angles. |
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| 413. |
The sum of either pair of opposite angles of a cyclic quadrilateral is `180^0`ORThe opposite angles of a cyclic quadrilateral are supplementary. |
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Answer» TO prove`=/_A+/_C=180^@,/_B+/_DD=180^@` Chord AB`/_ACB=/_ADB-(1)` Chord BC`/_BAC=/_BDC-(2)` adding equation 1 and 2 `/_ACB+/_BAC=/_ADB+/_BDC` `/_ACB+/_BAC+/_ABC=/_ADC+/_ABC` `/_D+/_B=180^@` `/_B+/_D=180^@` `/_A+/_B+/_C+/_D=360^@` `/_A+/_C+180^@=360^@=180^@` `/_B+/_D=180^@`. |
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| 414. |
In Figure, `P Q R S`is a cyclic quadrilateral. Find the measure of each of its angles. |
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Answer» `/_P+/_R=180^@` `/_B+/_S=180^@` `3x+x=180^@` `4x=180^@` `x=45^@` `/_P=3x=135^@` `/_R=x=45^@` `y+5y=180^@` `6y=180^@` `y=30^@` `/_B=y=30^@` `/_S=5y=150^@`. |
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| 415. |
A tangent PT is drawn to the circle `x^2 + y^2= 4` at the point `P(sqrt3,1)`. A straight line L is perpendicular to PT is a tangent to the circle `(x-3)^2 + y^2 = 1`Common tangent of two circle is: (A) `x=4` (B) `y=2` (C) `x+(sqrt3)y=4` (D) `x+2(sqrt2)y=6`A. `x-sqrt3y=1`B. `x+sqrt3y=1`C. `x-sqrt3y=-1`D. `x+sqrt3y=5` |
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Answer» Here, tangent to `x^(2)+y^(2)=4 "at" (sqrt3,1) "is" sqrt3x+y=4" "...(i)` AS, L is perpendicular to `sqrt3x + y = 4` `rArrx-sqrt3y=lambda` which is tangent to `(x-3)^(2)+y^(2)=1` `rArr (|3-0-lambda|)/(sqrt(1+3))=1` `rArr |3-lambda|=2` `rArr 3-lambda=2,-2` `therefore lambda=1,5` `rArrL:x-sqrt3y=1, x=-sqrt3y=5` |
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| 416. |
Tangents are drawn from the point (17, 7) to the circle `x^2+y^2=169`, Statement I The tangents are mutually perpendicular Statement, lls The locus of the points frorn which mutually perpendicular tangents can be drawn to the given circle is `x^2 +y^2=338` (a) Statement I is correct, Statement II is correct; Statement II is a correct explanation for Statementl (b( Statement I is correct, Statement I| is correct Statement II is not a correct explanation for Statementl (c)Statement I is correct, Statement II is incorrect (d) Statement I is incorrect, Statement II is correctA. Statement I is true, Statement II is true, Statement II is correct explanation of Statement I .B. Statement I is true, Statement II is true, Statement II is not correct explanation of Statement I.C. Statement I is true, Statement II is false.D. Statement I is false, Statement II is true. |
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Answer» Correct Answer - A As locus of point of intersection for perpendicular tangents is directors circle. i.e `x^(2)+y^(2)=2r^(2)` Hence, (17, 7) lie on directors circle `x^(2)+y^(2)`= 338 `rArr` tangents are perpendicular. |
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| 417. |
Consider a family of circles passing through the point (3,7) and (6,5). Answer the following questions. If the circle which belongs to the given family cuts the circle `x^(2)+y^(20=29` orthogonally, then the center of that circle isA. `(1//2,3//2)`B. `(9//2,7//2)`C. `(7//2,9//2)`D. `(3,-7//9)` |
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Answer» Correct Answer - 3 The equation of the line passing through the points A(3,7) and B(6,5) is `y-7= - (2)/(3) (x-3)` or `2x+3y -27 =0` Also, the equation of the circle with A and B as the endpoints of diameter is `(x-3) (x-6) +(y-7) (y-5) =0` Now, the equation of the family of circles through A and B is `(x-3)(x-6)+(y-7)(y-5)+lambda(2x+3y-27)=0` (1) If the circle belonging to this family touches the x-axis , then equation `(x-3)(x-6)+(0-7)(-5)+lambda { 2x+3(0)-27}=0` has two equal roots, for which discriminanat `D=0` . It gives two values of `lambda`. The equation of the common chord of (1) and `x^(2)+y^(2)-4x-6y-3=0` is the radical axis,which is `[(x-3)(x-6)+(y-7)(y-5)+lambda(2x+3y-27)] -[x^(2)+y^(2)-4x-6y-3]=0` or `(2 lambda -5)x+(3 lambda -6)y +(-27 lambda + 56) = 0` or `(-5x-6y+56)+lambda(2x+3y-27)=0` This is the family of lines which passes through the point of intersection of `-5x-6y+56=0` and `2x+3y-27=0, i.e., (2,23//3)`. If circle `(i)` cuts `x^(2)+y^(2)=29` orthogonally , then `0+0= -29 +56-27 lambda =0` or `lamda =1` So, the required circle is `x^(2)+y^(2)-7x-9y+26=0` and the center is `(7//2,9//2)`. |
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| 418. |
Find the equations of the circles passing through the point `(-4,3)`and touching the lines `x+y=2`and `x-y=2` |
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Answer» Correct Answer - `x^(2)+y^(2)+2(10pm3sqrt6)x+(55pm24sqrt6)=0` Let the equation of the required circle be `x^(2)+y^(2)+2gx+2fy+c=0" "...(i)` It passes through (-4, 3). `25-8g+6f+c=0" "...(ii)` Since, circle touches the line x + y - 2 = 0 and x - y - 2 = 0. `therefore |(-g-f-2)/(sqrt2)|=|(-g+f-2)/(sqrt2)|=sqrt(g^(2)+f^(2)-c)" "...(iii)` Now, ` |(-g-f-2)/(sqrt2)|=|(-g+f-2)/(sqrt2)|` `rArr -g-f-2=pm(-g+f-2)` `rArr -g-f-2=-g+f-2` or `rArr -g-f-2=g-f+2` `rArr f = 0 or g=-2` Case I when f = 0 From Eq. (iii) we get `|(-g-2)/(sqrt2)|=sqrt(g^(2)-c)` `rArr (g+2)^(2)= 2(g^(2)-c)` `rArr g^(2)-4g-4-2c=0" "...(iv)` On putting f = 0 in Eq. (ii) we get 25 = 8g + c = 0 ...(v) Eliminating c between Eqs. (iv) and (v),we get `g^(2)-20g+46=0` `rArr g = 10 pm3sqrt6 and c = 55pm24sqrt6` On substracting the value of g, f and c in Eq.(i) we get `x^(2)+y^(2)+2(10pm3sqrt6)x+(55pm24sqrt6)=0` Case II when g = - 2 From Eq. (iii), we get `rArr f^(2)=2(4+f^(2)-c)` `rArr f^(2)-2c + 8 = 0 " "...(vi)` On putting g = -2 in Eq. (ii), we get c = -6f - 41 ON substituting c in Eq.(vi), we get `f^(2)+12f+90=0` this equation gives imageinary value of F. Thus, there is no circle in this case. Hence, the required equations of the circles are `x^(2)+y^(2)+2(10pm3sqrt6)x+(55pm24sqrt6)=0` |
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| 419. |
Bisectors ofangles A, B and C of a triangle ABC intersect its circumcircle at D, E andFrespectively. Prove that the angles of the triangle DEF are `90o-1/2A`,`90o-1/2B`and `90o-1/2C` |
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Answer» `/_5=/_A/2` Since, angles in same segment at a circle are equal `/_6=/_B/2` `/_1=/_B/2` `/_2=/_C/2` `/_3=/_A/2` `/_4=/_C/2` `In/_DEF` `/_E=/_3+/_4=/_A/2+/_C/2=(/_A+/_C)/2` `=(180-/_B)/2` `/_E=90^@-/_B/2` `/_D=/_B/2+/_C/2=(/_B+/_C)/2` `/_D=90^@-/_A/2` `/_F=/_5+/_6` `=/_A/2+/_B/2=(/_A+/_B)/2` `/_F=90^@-/_C/2`. |
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| 420. |
`C_(1)` is a circle of radius 2 touching X-axis and Y-axis. `C_(2)` is another circle of radius greater than 2 and touching the axes as well as the circle `C_(1)` Statemnet I Radius of Circle `C_(2)=sqrt2(sqrt2+1)(sqrt2+2)` Statement II Centres of both circles always lie on the line y=x.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 - `:.` Statements I is true and Statements II is always not true (where circles in II of IV quadrants) |
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| 421. |
Prove thatthe angle in a segment greater than a semi-circle is less than a right angle. |
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Answer» According to diagram `180^@<2theta` `theta>90^@` `/_PBQ>90^@`. |
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| 422. |
Find the parametric form of the equation of the circle `x^2+y^2+p x+p y=0.` |
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Answer» The equation of the circle can be rewritten in the form `(x+(P)/(2))^(2)+(y+(p)/(2))^(2)=(p^(2))/(2)` Therefore, the parametric form of the equation of the given circle is `x=-(p)/(2)+(p)/(sqrt(2))cos theta` `=(p)/(2)(-1+sqrt(2)cos theta)` and `y= -(p)/(2)+(p)/(sqrt(2)) sin theta` `=(p)/(2)(-1+sqrt(2)sin theta)` where `0 lethetalt2pi` |
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| 423. |
Find the centre and radius of the circle whose parametric equation is `x= -1+2 cos theta , y= 3+2 sin theta`. |
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Answer» We have `(x+1)/(2)=cos theta` and `(y-3)/(2)=sin theta` Squaring and adding, we get `((x+1)/(2))^(2)+((y-3)/(2))^(2)=1` or `(x+1)^(2)+(y-3)^(2)=4` Centre of the circle is `(-1,3)` and radius is 2. |
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| 424. |
In given figure , ABC is a triangle produced meets the circumcircle of `Delta ABC` at `Q`, prove that `CP=CQ` |
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Answer» In ` Delta ABP`, `AB=AP` (given) `angle1=angle2` (angles opposite to equal sides ) .......(1) But `angle2 =angle3 ` (vertically opposite angle) ......(2) `rArrangle1=angle3` [From (1) and (2)] .........(3) But `angle1=angle4` (angles of same segment) .........(4) `therefore angle3= angle4` [from (3)and (4)] `rArrCP=CQ` (side opposite to equal of `DeltaCPQ`) |
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| 425. |
Find the equation of chord of the circle `x^(2)+y^(2)-2x-4y-4=0` passing through the point (2,3) which has shortest length. |
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Answer» Clearly required chord of the circle is that one which is bisected at point (2,3) . So, using `T=S_(1)` , equation of chord is `2x+3y-(x+2)-2(y+3)-4=2^(2)+3^(2)-2(2)-4(3)-4` or `x+y-5=0` |
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| 426. |
The chords of contact of tangents from three points `A ,Ba n dC`to the circle `x^2+y^2=a^2`are concurrent. Then `A ,Ba n dC`willbe concyclic(b) be collinearform the vertices of a trianglenone of theseA. be concyclicB. be collinearC. form the vertices of a triangleD. none of these |
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Answer» Correct Answer - 2 Let the coordinates of A,B, and C be `(x_(1),y_(1)),(x_(2),y_(2))` and `(x_(3),y_(3))` , respectively. Then, the chords of contact of tangents drawn from A,B, and C are `x x_(1)+y y_(1)= a^(2), x x_(2)+y y_(2) =a^(2), ` and `x x _(3)+y y_(3)=a^(2)`, respectively. These three lines will be concurrent , if `|{:(x_(1),y_(1),-a^(2)),(x_(2),y_(2),-a^(2)),(x_(3),y_(3),-a^(2)):}|=0` or `|{:(x_(1),y_(1),1),(x_(2),y_(2),1),(x_(3),y_(3),1):}|=0` Therefore, points `(x_(1),y_(1)),(x_(2),y_(2))` and `(x_(3),y_(3))` are collinear. |
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| 427. |
The equation of the chord of the circle `x^2+y^2-3x-4y-4=0`, which passes through the origin such that the origin divides it inthe ratio 4:1, is`x=0`(b) `24 x+7y=0``7x+24=0`(d) `7x-24 y=0`A. `x=0`B. `24x+7y=0`C. `7x+24y=0`D. `7x-24y=0` |
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Answer» Correct Answer - 2 Let `y=mx` be a chord. Then the points of intersections are given by `x^(2)(1+m^(2))-x(3+4m)-4=0` `:. x_(1)+x_(2)=(3+4m)/(1+m^(2))` and `x_(1)x_(2)=(-4)/(1+m^(2))` Since `(0,0)` divides chord in the ratio `1:4`, we have `x_(2)= -4x_(1)` `:. -3x_(1)=(3+4m)/(1+m^(2))` and `4x_(1)^(2)=(-4)/(1+m^(2))` `:. 9+9m^(2)=9=9+16m^(2)+24m` i.e., `m=0,-(24)/(7)` Therefore, the lines are `y=0` and `y+24x=0` |
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| 428. |
If `O Aa n dO B`are equal perpendicular chordsof the circles `x^2+y^2-2x+4y=0`, then the equations of `O Aa n dO B`are, where `O`is the origin.`3x+y=0`and `3x-y=0``3x+y=0`and `3y-x=0``x+3y=0`and `y-3x=0``x+y=0`and `x-y=0`A. `3x+y=0` and `3x-y=0`B. `3x+y=0` and `3y-x=0`C. `x+3y=0` and `y-3x=0`D. `x+y=0` and `x-y=0` |
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Answer» Correct Answer - 3 Let the equation of the chord OA of the circle `x^(2)+y^(2)-2x+4y=0` (1) by `y=mx` (2) Solving (1) and (2) , we get `x^(2)+m^(2)x^(2)-2x+4x=0` or `(1+m^(2))x^(2)-(2-4m)x=0` or `x=0` and `x=(2-4m)/(1+m^(2))` Hence, the points of intersection are `O(0,0)` and `A((2-4m)/(1+m^(2)),(m(2-4m))/(1+m^(2)))` or `OA^(2)=((2-4m)/(1+m^(2)))(1+m^(2))=((2-4m)^(2))/(1+m^(2))` Since OAB is an isosceles right-angles triagnle, `OA^(2)=(1)/(2)AB^(2)` where AB is a diameter of the given circle. Hence, `OA^(2)=10` or `((2-4m)^(2))/(1+m^(2))=10` or `4-16m+16m^(2)=10+10m^(2)` or `3m^(2)-8m-3=0` i.e., `m=3` or `-(1)/(3)` Hence, the required equations are `y=3x` or `x+3y=0` |
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| 429. |
If `C_1,C_2,a n dC_3`belong to a family of circles through the points `(x_1,y_2)a n d(x_2, y_2)`prove that the ratio of the length of the tangents from any point on `C_1`to the circles `C_2a n dC_3`is constant. |
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Answer» Equations of circles `C_(1),C_(2)` and `C_(3)` through A`(x_(1),y_(1))` and `B(x_(2),y_(2))` are given by ` ubrace((x-x_(1))(x-x_(2))+(y-y_(1))(y-y_(2)))_("f(x,y)")` ` +lambda_(r)ubrace((y-y_(1)-(y_(2)-y_(1))/ (x_(2)-x_(2))(x-x_(1))))_("g(x,y)")=0,r=1,2,3` or `f(x,y)+lambda_(r)g(x,y)=0,r=1,2,3` Consider point P(h,k) on circle `C_(1)`. `:. f(h,k)+lambda_(1)g(h,k)=0` (1) Let `T_(2)` and `T_(3)` be the lengths of the tangents from P to `C_(2)` and `C_(3)`, respectively. `:. (T_(2))/(T_(3))=(sqrt(f(h,k)+lambda_(2)g(h,k)))/(sqrt(f(h,k)+lambda_(3)g(h,k)))` `=(sqrt(-lambda_(1)g(h,k)+lambda_(2)g(h,k)))/(sqrt(-lambda_(1)g(h,k)+lambda_(3)(h,k)))=sqrt((lambda_(2)-lambda_(1))/(lambda_(3)-lambda_(1)))` Clearly, this ratio is independtn of the choice of (h,k) and hence, a constant. |
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| 430. |
Find the number of common tangent to the circles `x^2+y^2+2x+8y-23=0`and `x^2+y^2-4x-10 y+9=0`A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - 3 `C_(1)-=(-1,-4),C_(2) -=(2,5)` `r_(1)=sqrt(1+16+23)=2sqrt(10)` `r_(2)=sqrt(4+25-19)=sqrt(10)` `C_(1)C_(2)=sqrt(9+81)=3sqrt(10)` `:. C_(1)C_(2)=r_(1)+r_(2)` Hence, the circle touch externally. |
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| 431. |
A is a point (a, b) in the first quadrant. If the two circIes which passes through A and touches the coordinate axes cut at right angles then :A. `a^(2)-6ab+b^(2)=0`B. `a^(2)+2ab-b^(2)=0`C. `a^(2)-4ab+b^(2)=0`D. `a^(2)-8ab+b^(2)=0` |
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Answer» Correct Answer - 3 Let the equation of the two circles be `(x-r)^(2)+(y-r)^(2)=r^(2)` i.e.., `x^(2)+y^(2)-2rx-2ry+r^(2)=0`, where `r=r_(1)r_(2)` The condition of orthogonality gives `2r_(1)r_(2)+2r_(1)r_(2)=r_(1)^(2)+r_(2)^(2)` or `4r_(1)r_(2)=r_(1)^(2)=r_(2)^(2)` (1) Circle passes through `(a,b)`. Therefore, `a^(2)+b^(2)-2ra-2rb+r^(2)=0` i.e., `r^(2)-2r(a+b)+a^(2)+b^(2)=0` `:. r_(2) +r_(2)=2(a+b)` and `r_(1)r_(2)=a^(2)+b^(2)` `:. 4(a^(2)+b^(2))=4(a+b)^(2)-2(a^(2)+b^(2))` [From (1)] i.e., `a^(2)-4ab+b^(2)=0` |
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| 432. |
In the figure, M is the centre of the circle and seg KL is a tangent segment. If `MK = 12 , KL = 6 sqrt(3)` then find, (1) Radius of the circle, (2) Measures of `/_ K ` and `/_ M `. |
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Answer» In `Delta MLK`, `/_ MLK = 90^(@)` ….(Tangent theorem) `:.` by Pythagoras theorem, `MK^(2) = ML^(2) + LK^(2)` `:. 12^(2) = ML^(2) + ( 6 sqrt(3))^(2) ` `:. 144 = ML^(2) + 36 xx 3 ` `:. 144 = ML^(2) + 108` `:. ML^(2) = 144 - 108` `:. ML^(2) = 36` `:. ML= 6` ...(Takin quare roots of both the sides ) `:. ` radius of the circle = ML = 6 . In `Delta MLK`, `ML = (1)/(2) MK ` `:. /_ K = 30^(@)` ...(By converse of `30^(@) - 60^(@) - 90^(@)` triangle theorem ) In `Delta MLK`, `/_ M + /_ K + /_ L = 180^(@)` ...(Sum of all angles of a triangle is `180^(@)0` `:. /_ M + 30^(@) + 90^(@) = 180^(@0` `:. /_ M + 120^(@) = 180^(@)` `:. /_ M = 180^(@) - 120^(@)` `:. /_ M = 60^(@)` |
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| 433. |
If the circle `x^2+y^2+2gx+2fy+c=0`is touched by `y=x`at `P`such that `O P=6sqrt(2),`then the value of `c`is36 (b) 144(c) 72 (d)none of theseA. 36B. 144C. 72D. none of these |
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Answer» Correct Answer - 3 The equation of the line `y=x` in distance form is `(x)/(cos theta)=(y)/(sin theta)=r`, where ` theta =(pi)/(4)` For point P`, r=6 sqrt(2)`. Therefore, the coordinates of P are given by `(x)/(cos (pi//4))=(y)/(sin (pi//4))=6 sqrt(2)` or `x=6,y=6` Since `y=x` touches the circle, the equation `2x^(2)+2x(g+f)+c=0` has the equal roots . Therefore, `4(g+f)^(2)=8c` or `(g+f)^(2)=2c` (2) From (1), we get `[12(g+f)^(2)=[-(c+72)]^(2)` or `144(g+f)^(2)=(c+72)` or`144(2c)=(c+72)^(2)` or `(c-72)^(2)=0` or `c=72` |
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| 434. |
The equation of the circumcircle of an equilateral triangle is `x^2+y^2+2gx+2fy+c=0`and one vertex of the triangle in (1, 1). The equation of the incircleof the triangle is`4(x^2+y^2)=g^2+f^2``4(x^2+y^2)=8gx+8fy=(1-g)(1+3g)+(1-f)(1+3f)``4(x^2+y^2)=8gx+8fy=g^2+f^2``non eoft h e s e`A. `4(x^(2)+y^(2))=g^(2)+f^(2)`B. `4(x^(2)+y^(2))+8gx+8fy=(1-g)(1+3g)+(1-f)(1+3f)`C. `4(x^(2)+y^(2))+8gx+8fy=g^(2)+f^(2)`D. none of these |
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Answer» Correct Answer - 2 In an equilateral triangle, circumcenter and incenter are coincident. Therefore, Inceter`-= ( -g,-f)` Point (1,1) lies on the circle. Therefore, `1^(2)+1^(2)+2g+2f+c=0` or `c= -2(g+f+1)` Also, in an equilateral triangle, Circumradius`=2xx` Inradius `:. `Inradius `=(1)/(2) xx sqrt(g^(2)+f^(2)-c)` Therefore, the equation of the incircle is `(x+g)^(2)+(y+f)^(2)=(1)/(4)(g^(2)+f^(2)-c)` ,brgt `=(1)/(4){g^(2)+f^(2)+2(g+f+1}` |
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| 435. |
A variable circle C has the equation `x^2 + y^2 - 2(t^2 - 3t+1)x - 2(t^2 + 2t)y + t = 0`, where t is a parameter.The locus of the centre of the circle isA. `((1)/(10),-(1)/(10))`B. `((1)/(10),(1)/(10))`C. `(-(1)/(10),(1)/(10))`D. `(-(1)/(10),-(1)/(10))` |
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Answer» Correct Answer - Hence, required ordered pair is `(-(1)/(10),(1)/(10))` |
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