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InterviewSolution
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.
| 1. |
The loucs of the centre of the circle which cuts orthogonally the circle `x^(2)+y^(2)-20x+4=0` and which touches x=2 isA. `x^(2)=16y`B. `x^(2)=16y+4`C. `y^(2)=16x`D. `y^(2)=16x+4` |
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Answer» Correct Answer - C |
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| 2. |
If a variable line, `3x + 4y -lambda = 0` is such that the two circles `x^(2)+y^(2)-2x-2y+1=0` and `x^(2)+y^(2)-18x-2y+78=0` are not its opposite sides, then the set of all values of `lambda` is the intervalA. [13, 23]B. [2,17]C. [12,21]D. [23,31] |
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Answer» Correct Answer - C The given circle, `x^(2)+y^(2)-2x-2y+1=0" "...(i)` and `x^(2)+y^(2)-18x-2y+78=0," "...(ii)` are on the opposite side of the variable line `3x+4y-lambda = 0`. Their centres also lie in the opposite sides of the varible line. `rArr[3(1)+4(1)-lambda][3(9)+4(1)-lambda]lt0` [`therefore ` the points ` P(x_(1), y_(1)) and Q (x_(2), y_(2))` lie on the opposite sides of the line ax + by + c =0, `if (ax_(1)+by_(1)+c)(ax_(2)+by_(2)+c)lt0]` `rArr (lambda-7)(lambda-31)lt0` `rArr lambdain(7, 31)" "...(iii)` Also, we have `|(3(1)+4(1)-lambda)/(5)|gesqrt(1+1-1)` `(therefore" Distance of centre from the given line is greater than the radius, i.e., " (ax_(1)+by_(1)+c)/(sqrt(a^(2)+b^(2)))ge r)` `rArr |7-lambda| ge5rArrlambdain(-oo,2] uu [12,oo)" "(iv)` and `|(3(9)+4(1)-lambda)/(5)|gesqrt((81+1-78))` `rArr|lambda-31|ge10` `rArr lambdain(-oo,21]uu[,oo)" "...(v)` From Eqs. (iii), (iv) and (v), we get `lambda in [12, 21]` |
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| 3. |
The equation of a circle which cuts the three circles `x^(2)+y^(2)+2x+4y+1=0,x^(2)+y^(2)-x-4y+8=0` and `x^(2)+y^(2)+2x-6y+9=0` orthogonlly isA. `x^(2)+y^(2)-2x-4y+1=0`B. `x^(2)+y^(2)+2x+4y+1=0`C. `x^(2)+y^(2)-2x+4y+1=0`D. `x^(2)+y^(2)-2x-4y-1=0` |
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Answer» Correct Answer - A |
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| 4. |
Two variable chords AB and BC of a circle `x^(2)+y^(2)=a^(2)` are such that `AB=BC=a`. M and N are the midpoints of AB and BC, respectively, such that the line joining MN intersects the circles at P and Q, where P is closer to AB and O is the center of the circle. The locus of the points of intersection of tangents at A and C isA. `x^(2)+y^(2)=a^(2)`B. `x^(2)+y^(2)=2a^(2)`C. `x^(2)+y^(2)=4a^(2)`D. `x^(2)+y^(2)=8a^(2)` |
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Answer» Correct Answer - C |
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| 5. |
One of the limiting points of the co-axial system of circles containing the circles `x^(2)+y^(2)-4=0andx^(2)+y^(2)-x-y=0` isA. `(sqrt2,sqrt2)`B. `(-sqrt2,sqrt2)`C. `(-sqrt2-sqrt2)`D. None of these |
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Answer» Correct Answer - D |
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| 6. |
The limiting points of the system of circles represented by the equation `2(x^(2)+y^(2))+lambda x+(9)/(2)=0`, areA. `(pm(3)/(2),0)`B. `(0,0)and((9)/(2),0)`C. `(pm(9)/(2),0)`D. `(pm3,0)` |
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Answer» Correct Answer - A |
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| 7. |
`t_(1),t_(2),t_(3)` are lengths of tangents drawn from a point (h,k) to the circles `x^(2)+y^(2)=4,x^(2)+y^(2)-4=0andx^(2)+y^(2)-4y=0` respectively further, `t_(1)^(4)=t_(2)^(2)" "t_(3)^(2)+16`. Locus of the point (h,k) consist of a straight line `L_(1)` and a circle `C_(1)` passing through origin. A circle `C_(2)` , which is equal to circle `C_(1)` is drawn touching the line `L_(1)` and the circle `C_(1)` externally. Equation of `L_(1)` isA. x+y=0B. x-y=0C. 2x+y=0D. x+2y=0 |
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Answer» Correct Answer - A |
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| 8. |
Find the equation of the radical axis of circles `x^2+y^2+x-y+2=0` and `3x^2+3y^2-4x-12=0`A. `2x^(2)+2y^(2)-5x+y-14=0`B. `7x-3y+18=0`C. `5x-y+14=0`D. None of these |
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Answer» Correct Answer - B |
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| 9. |
Find the equation of radical axis of the circles `x^(2)+y^(2)-3x+5y-7=0` and `2x^(2)+2y^(2)-4x+8y-13=0`. |
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Answer» We have circles `x^(2)+y^(2)-3x+5y-7=0` (1) and `2x^(2)+2y^(2)-4x+8y-13=0` (2) Equation of radical axis of the two circles can be obtained by subtracting the equation of one circle from that of the other in such a way that term `x^(2)` and `y^(2)` gets cancelled out. So, multiplying equation (1) by 2 and then subtracting equation (2) from it, we get ,brgt `-2x+2y-1=0` or `2x-2y+1=0` |
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| 10. |
The radius centre of the circles `x^(2)+y^(2)=1,x^(2)+y^(2)+10y+24=0andx^(2)+y^(2)-8x+15=0` isA. (2,5/2)B. (-2,5/2)C. (-2,-5/2)D. (2,-5/2) |
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Answer» Correct Answer - D |
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| 11. |
The equation of three circles are given `x^2+y^2=1,x^2+y^2-8x+15=0,x^2+y^2+10 y+24=0`. Determine the coordinates of the point `P`such that the tangents drawn from it to the circle are equal in length. |
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Answer» We known that the point from which the lengths of tangents are equal in length is the radical of the given three circles. Now, the radical axis of the first two circles is `(x^(2)+y^(2)-1)-(x^(2)+y^(2)-8x+15)=0` `i.e., x-2=0` (1) and the radical axis of the second and third circle is `(x^(2)+y^(2)-8x+15)-(x^(2)+y^(2)+10y+24)=0` i.e., `8x+10y+9=0` (2) Solving (1) and (2), the coordinates of the radical center, i.e., of the point P, are `P(2,-5//2)`. |
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| 12. |
If two circles `x^(2)+y^(2)+c^(2)=2ax` and `x^(2)+y^(2)+c^(2)-2by=0` touch each other externally , then prove that `(1)/(a^(2))+(1)/(b^(2))=(1)/(c^(2))` |
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Answer» The two circles are `x^(2)+y^(2)-2ax+c^(2)=0` and `x^(2)+y^(2)-2by+c^(2)=0`. Respective centres are `C_(1)(a,0)` and `C_(2)(0,b)`. Respective radii are `r_(1)=sqrt(a^(2)-c^(2))` and `r_(2)=sqrt(b^(2)-c^(2))` Since the two circles touch each other externally , we have `C_(1)C_(2)=r_(1)+r_(2)`. `implies sqrt(a^(2)+b^(2))=sqrt(a^(2)-c^(2))+sqrt(b^(2)-c^(2))` `implies a^(2)+b^(2)=a^(2)-c^(2)+b^(2)-c^(2)+2sqrt(a^(2)-c^(2))sqrt(b^(2)-c^(2))` `implies c^(2)=sqrt(a^(2)-c^(2))sqrt(b^(2)-c^(2))` `implies c^(4)=a^(2)b^(2)-c^(2)(a^(2)+b^(2))+c^(4)` `implies a^(2)b^(2)=c^(2)(a^(2)+b^(2))` `implies (1)/(a^(2))+(1)/(b^(2))=(1)/(c^(2))` |
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| 13. |
Show that the circles `x^2+y^2-10 x+4y-20=0`and `x^2+y^2+14 x-6y+22=0`touch each other. Find the coordinates of the point of contact and theequation of the common tangent at the point of contact. |
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Answer» Two circles are `S_(1)=x^(2)+y^(2)-10x+4y-20=0` (1) and `S_(2)=x^(2)+y^(2)+14x-6y+22=0` (2) For the circle (1), centre is `C_(1)(5,-2)` and radius, `r_(1)=sqrt((-5)^(2)+2^(2)-(-20))=7`. For the circle (2), centre is `C_(2)(7,-3)` and radius, `r_(2)=sqrt(7^(2)+(-3)^(2)-22)=6` Now, `C_(1)C_(2)=sqrt((5+7)^(2)+(-2,3)^(2))=13=r_(1)+r_(2)` Thus, two circle are touching each other externally. Point of contact T divides `C_(1)C_(20` internally in the ration `r_(1) : r_(2)=7:6`. `:. T=((7(-7)+6(5))/(7+6),(7(3)+6(-2))/(7+6))-=(-19//3,9//13)` Common tangents at point of contact is radical axis, whose equation is given by `S_(1)-S_(2)=0`. Therefore, equation of common tangent is `-24x+10y-42=0` or `12x-5y+21=0`. |
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| 14. |
The number of intergral value of `y`for which the chord of the circle `x^2+y^2=125`passing through the point `P(8,y)`gets bisected at the point `P(8,y)`and has integral slope is8 (b) 6(c) 4 (d)2A. 8B. 6C. 4D. 2 |
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Answer» Correct Answer - 2 The slope of the chord is `m= -8//y`. Therefore, `y= +-1,+-2,+-4,+-8` But (8,y) must also lie inside the circle `x^(2)+y^(2)=125` So, y can be equal to `+-1,+-2,+-4,` i.e., 6 value |
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| 15. |
Find the coordinates of the point at which the circles `x^2-y^2-4x-2y+4=0` and `x^2+y^2-12 x-8y+36=0` touch each other. Also, find equations of common tangents touching the circles the distinct points. |
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Answer» Correct Answer - `y = 0 and 7y -24x + 16=0` two circles touch each other externally, if `C_(1)C_(2) = r_(1)+r_(2)` and internally if `C_(1)C_(2)= r_(1)~r_(2)` Given circles are `x^(2)+y^(2) -2y+4=0`, whose centre `C_(1)(2, 1)` and radius `r_(1)=1` and `x^(2) +y^(2)- 12x -8y +36=0` whose centre `C_(2)(6,4)` and radius `r_(2)=4` The distance between the centres is `sqrt((6-2)^(2)+(4-1)^(2))=sqrt(16+9)=5` `rArr C_(1) C_(2) =r_(1)+r_(2)` Therefore, the circles touch each other externally and at the point of touching the point divides the line joining the two centres internally in the ratio of their radii, `1 : 4` Therefore, `x_(1) = (1xx6+4xx2)/(1+4)=(14)/(5)` `y_(1)= (1xx4+4xx1)/(1+4)=(8)/(5)` Again, to determine the equation of common tangents touching the circles in distinct points, we know that, the tangents pass through a point which divides the line joining the two centres externally in the ratio of their radi, i.e. `1:4` Therefore, `x_(2)=(1xx6-4xx2)/(1-4)=(-2)/(-3)=(2)/(3)` `and y_(2)=(1xx4-4xx1)/(1-4)=0` Now, let m be the slope of the tangent and this line passing through (2/3, 0 ) is `y-0=m(x-2//3)` `y-mx+(2)/(3)m=0` This is tangent to the Ist circle, if perpendicular distance from centre = radius. `therefore(1-2m+(2//3)m)/(sqrt(1+m^(2)))=1 [therefore C_(1)=(2,1) and r_(1)=1]` `rArr 1-(4)/(3)m=sqrt(1+m^(2))` `rArr1+(16)/(9)m^(2)-(8)/(3)m=1+m^(2)` `rArr (7)/(3)m^(2)-(8)/(3)m=0` `rArr m((7)/(9)m-(8)/3)=0` `rArr m = 0 ,m=(27)/(7)` Hence, the equation of two tangents are `y=0and y=(27)/(7)(x-(2)/(3))` `rArr y = 0 and 7y -24x + 16 = 0` |
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| 16. |
Find the center of the smallest circle which cuts circles `x^2+y^2=1`and `x^2+y^2+8x+8y-33=0`orthogonally. |
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Answer» Let the equation of the required circle be `x^(2)+y^(2)+2gx+2fy+c=0`. This circle cuts the given two circles othogonally. `:. 2g(0)+2f(0)= -1+c,` `:. c=1` and `2(g)(4)+2(f)(4)= -33+c= -32` `:. G+f= -4` `:. `Radius of the circle `sqrt(g^(2)+f^(2)-c)` `=sqrt(g^(2)+(g+4)^(2)-1) ` `=sqrt(2g^(2)+8g+15)` `=sqrt(2(g+2)^(2)+7)` Radius is minimum if `g+2=0` or `g =-2`. `:. f= -1` Hence, equation of the circle is `x^(2)+y^(2)-4x-4y+1=0`. |
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| 17. |
In the figure, ray PQ touches the circle at point Q. If PQ =12, PR = 8, then find PS. |
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Answer» Correct Answer - PS = 18 Ray PQ is tangent touching the circle at point Q and line PRS is secant intersecting the circle at points R and S. `:. PQ^(2) = PR xx PS ` …( By tangent secant segments theorem ) `:. 12^(2) = 8 xx PS ` `:. PS = (12 xx 12)/( 8)` `:. PS = 18` |
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| 18. |
The distance from the center of the circle `x^2+y^2=2x`to the common chord of the circles `x^2+y^2+5x-8y+1=0`and `x^2+y^2-3x+7y-25=0`is2 (b) 4(c) `(34)/(13)`(d) `(26)/(17)`A. 2B. 4C. `34//13`D. `26//17` |
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Answer» Correct Answer - 1 The center of the circle `x^(2)+y^(2)=2x` is (1,0). The common chord of the other two circle is `8x-15y+26=0` The distance from (1,0) to `8x-15y+26=0` is `(|8+26|)/(sqrt(15^(2)+8^(2)))=2` |
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| 19. |
From the point A `(0,3)` on the circle `x^(2)+ 4x+(y-3)^(2) = 0` a chord AB is drawn and extended to a point M such that AM = 2 AB. Find the equation of the locus of M.A.B.C.D. |
| Answer» Correct Answer - `x^(2)+ y^(2) + 8x - 6y + 9 = 0` | |
| 20. |
A circle S passes through the point (0, 1) and is orthogonal to the circles `(x -1)^2 + y^2 = 16` and `x^2 + y^2 = 1`. Then(A) radius of S is 8(B) radius of S is 7(C) center of S is (-7,1)(D) center of S is (-8,1)A. radius of S is 8B. radius of S is 7C. centre of S is `( -7,1)`D. centre of S is `( -8,1)` |
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Answer» Correct Answer - 2,3 Given circles `S_(1) : x^(2)+y^(2)-2x-15=0` and `S_(2) : x^(2)+y^(2)-1=0` Center of the circle which intersects above two circles orthogonally lies on the radical axis of the circles which is `S_(1)-S_(2) =0` or `x+7=0` Let the centre of the required circle be `C (-7,k)` Circle passes throught the point `A (0,1)`. `:. ` radius , `sqrt(7^(2)+(k-1)^(2))` Also, radius `=` length of the tangent from C to the any one of the given circles. `:. r=sqrt(7^(2)+k^(2)-1)` Comparing , we get `7^(2)+(k-1)^(2)=7^(2)+k^(2)-1` or `-2k+1= -1` or `k=1` `:. r=7` |
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| 21. |
A circle S passes through the point (0, 1) and is orthogonal to the circles `(x -1)^2 + y^2 = 16` and `x^2 + y^2 = 1`. Then(A) radius of S is 8(B) radius of S is 7(C) center of S is (-7,1)(D) center of S is (-8,1)A. radius of S is 8B. radius of S is 7C. centre of S is (-7,1)D. centre of S is (-8,1) |
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Answer» Correct Answer - B::C |
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| 22. |
If a line segement `A M=a`moves in the plane `X O Y`remaining parallel to `O X`so that the left endpoint `A`slides along the circle `x^2+y^2=a^2,`then the locus of `Mdot`A. `x^(2)+y^(2)=4a^(2)`B. `x^(2)+y^(2)=2ax`C. `x^(2)+y^(2)=2ay`D. `x^(2)+y^(2)-2ax-2ay=0` |
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Answer» Correct Answer - B |
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| 23. |
The point diametrically opposite to the point P(1, 0) on the circle `x^(2)+y^(2)+2x+4y-3=0` isA. (3,-4)B. (-3,4)C. (-3,-4)D. (3,4) |
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Answer» Correct Answer - C |
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| 24. |
If `alpha-` chord of a circle be that chord which subtends an angle `alpha` at the centre of the circle. If x+y=1 is `alpha`-chord of `x^(2)+y^(2)=1`, then `alpha` is equal toA. `(pi)/(6)`B. `(pi)/(4)`C. `(pi)/(2)`D. `(3pi)/(4)` |
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Answer» Correct Answer - `alpha=(pi)/(2)` |
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| 25. |
Consider the relation `4l^(2)-5m^(2)+6l+1=0`, where l, m `inR`. The line lx+my+1=0 touches a fixed circle whose equation isA. `x^(2)+y^(2)-4x-5=0`B. `x^(2)+y^(2)+6x+6=0`C. `x^(2)+y^(2)-6x+4=0`D. `x^(2)+y^(2)+4x-4=0` |
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Answer» Correct Answer - `x^(2)+y^(2)-6x+4=0` |
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| 26. |
Consider the relation `4l^(2)-5m^(2)+6l+1=0` , where `l,m in R` The number of tangents which can be drawn from the point (2,-3) to the above fixed circle are |
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Answer» Correct Answer - Therefore, point (2,-3) lies outside the circle from which two tangents can drawn. |
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| 27. |
Consider the relation `4l^(2)-5m^(2)+6l+1=0` , where `l,m in R` The number of tangents which can be drawn from the point (2,-3) to the above fixed circle areA. `((1)/(2),(5)/(2))`B. `((1)/(3),(4)/(3))`C. `(-(1)/(2),(3)/(2))`D. `((1)/(2),(5)/(2))` |
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Answer» Correct Answer - `:.` Fixed piont is `((1)/(2),(-5)/(2))` |
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| 28. |
If a circle is concentric with the circle `x^(2)+y^(2)-4x-6y+9=0` and passes through the point (-4,-5) then its equation isA. `x^(2)+y^(2)+4x+6y-87=0`B. `x^(2)+y^(2)+4x+6y+87=0`C. `x^(2)+y^(2)-4x-6y-87=0`D. `x^(2)+y^(2)-4x-6y+87=0` |
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Answer» Correct Answer - C |
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| 29. |
Find the equation of the circle concentric with the circle `x^2 + y^2 - 4x - 6y - 3 = 0` and which touches the y axis |
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Answer» Correct Answer - ` x ^(2) + y ^(2)- 4x - 6y + 9 = 0 ` Centre of the given circle is ` (2, 3 )` Centre of the required circle is ` (2, 3 )` The required circle touches the y- axis, so its radius = 2. `therefore ` the required equation is ` (x - 2 ) ^(2) + ( y- 3 )^(2) = 2 ^(2)`. |
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| 30. |
Find the equation of the circle concentric with the circle ` x ^(2) + y ^(2) + 4 x + 6y + 11 = 0 ` and passing through the point ` P(5, 4 )`. |
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Answer» Correct Answer - ` x ^(2) + y ^(2) + 4x + 6y - 85 = 0 ` Centre of the required circle is `C(-2, - 3 )`. The circle passes through the point ` P (5, 4 )` Radius ` = | CP| = sqrt(( 5= 2 ) ^(2) + ( 4+ 3) ^(2)) = 7sqrt 2 ` `therefore ` the required equation is ` (x + 2 ) ^(2) + ( y + 3) ^(2() = 98 ` |
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| 31. |
Show that the equation ` x ^(2) + y ^(2) - 3x + 3y + 10 = 0 ` does not represent a circle. |
| Answer» Show that ` (g ^(2) + f ^(2) - c ) lt 0 `, where ` g = ( -3)/(2) , f = (3)/(2) and c = 10`. | |
| 32. |
Find the equation of the circle which is circumscribed about the triangle whose vertices are ` A(- 2, 3 ) , B( 5, 2 ) and C (6, -1 )`. Find the centre and radius of this circle. |
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Answer» Correct Answer - ` x ^(2) + y^(2) - 2x + 2y - 23 = 0 `, centre `(1, - 1)` and radius = 5 Find the equation of the circle passing through the points `A(-2, 3), B(5, 2 ) and C(6, -1)`. |
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| 33. |
If the area of the quadrilateral by the tangents from the origin to thecircle `x^2+y^2+6x-10 y+c=0`and the radii corresponding to the points of contact is `15 ,`then a value of `c`is9 (b) 4(c) 5 (d) 25A. 9B. 4C. 5D. 25 |
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Answer» Correct Answer - 1,4 Area of quadrilateral `=sqrt(c ) xx sqrt(9+25 - c) =15` `:. c=9, 25` |
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| 34. |
Which of the following lines have the intercepts of equal lengths on the circle, `x^2+y^2-2x+4y=0`A. `3x-y=0`B. `x+3y=0`C. `x+3y+10=0`D. `3x-y-10=0` |
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Answer» Correct Answer - 1,2,3,4 Chords equidistant from the center are equal. |
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| 35. |
If the two circles `(x+1)^2+(y-3)=r^2 and x^2+y^2-8x+2y+8=0` intersect in two distinct point,then (A) `r > 2` (B) `2 < r < 8` (C) `r < 2` (D) `r=2` |
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Answer» We have circle`(x-1)^(2)+(y-3)^(2)=r^(2)` having centre `C_(1)(1,3)` and radius `r_(1)=r` and circle `x^(2)+^(2)-8x+2y+8=0` having centre `C_(2)(4,-1)` and `r_(2)=sqrt(4^(2)+(-1)^(2)-8)=3`. Circles intersect in two distinct points if `|r_(1)-r_(2)|ltC_(1)C_(2)ltr_(1)+r_(2)` `:. |r-3| lt 5ltr+3` `implies |r-3| lt5` and `rgt2` `implies -5 lt r-3lt5` and `rgt2` `implies -2 lt rlt8` and `rgt2` `implies 2ltrlt8` |
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| 36. |
If the two circles `(x+1)^2+(y-3)=r^2 and x^2+y^2-8x+2y+8=0` intersect in two distinct point,then (A) `r > 2` (B) `2 < r < 8` (C) `r < 2` (D) `r=2`A. `2ltrlt8`B. `rlt2`C. r = 2D. `r gt 2` |
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Answer» Correct Answer - A As, the two circle intersect in two distinct points. `rArr " Distance between centre lies between " |r_(1)-r_(2)| and |r_(1)+r_(2)|`. i.e. `|r-3|ltsqrt((4-1)^(2)+(-1-3)^(2))lt|r+3|` `rArr |r-3|lt5lt|r+3|rArrlt8 or rgt2` `therefore 2ltrlt8` |
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| 37. |
Let `0 lt alpha lt (pi)/(2) ` be a fixed angle . If `p=(costheta, sin theta) and Q(cos(alpha-theta))`, then Q is obtained from P byA. clockwise rotation around origin through an angle `alpha`B. anit-clockwise rotation around origin through an angle `alpha`C. reflection in the line through origin with slope `tanalpha`D. reflection in the line through origin which slope `tan((alpha)/(2))` |
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Answer» Correct Answer - D |
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| 38. |
S(x,y)=0 represents a circle. The equation S(x,2)=0 gives two identical solutions x=1 and the equation S(1,y)=0 gives two distinct solutions y=0,2 then the equatino of the circle isA. `x^(2)+y^(2)+2x-2y+1=0`B. `x^(2)+y^(2)-2x+2y+1=0`C. `x^(2)+y^(2)-2x-2y-1=0`D. `x^(2)+y^(2)-2x-2y+1=0` |
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Answer» Correct Answer - D |
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| 39. |
If two perpendicular tangents can be drawn from the origin to thecircle `x^2-6x+y^2-2p y+17=0`, then the value of `|p|`is___ |
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Answer» Correct Answer - 5 The equation of the given circle is `x^(2)+y^(2)-6x-2py+17=0` or `(x-3)^(2)+(u-p)^(2)=(p^(2)-8)` (1) Also, `(0,0)` lies outside the circle. The equation of director circle of `S=0` will be `(x-3)^(2)+(u-p)^(2)=2(p^(2)-8)` (2) Tangents drawn from `(0,0)` to circle (1) are perpendicular to each other. Therefore, `(0,0)` must lie on director circle. Hence, `(0-3)^(2)+(0-p)^(2)=2(p^(2)-8)` or `p^(2)=25` or `p= +-5` |
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| 40. |
Find the condition that the circle `(x-3)^2+(y-4)^2=r^2`lies entirely within the circle `x^2+y^2=R^2`.A. `R+rle7`B. `R^(2)+r^(2)lt49`C. `R^(2)-r^(2)lt25`D. `R-rgt5` |
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Answer» Correct Answer - D |
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| 41. |
If one of the circles `x^2+y^2+2ax+c=0` and `x^2+y^2+2bx+c=0` lies within the other, thenA. `abgt0,cgt0`B. `abgt0,clt0`C. `ablt0,cgt0`D. `ablt0,clt0` |
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Answer» Correct Answer - A |
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| 42. |
If real numbers `xa n dy`satisfy `(x+5)^2+(y-12)^2=(14)^2,`then the minimum value of `sqrt(x^2+y^2)`is_________ |
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Answer» Correct Answer - 1 Let `x+5=14 cos theta` and `y-12=14 sin theta`. Then `x^(2)+y^(2) = (14 cos theta - 5)^(2) + (14 sin theta +12)^(2)` `=196+25+144+28(12 sin theta - 5 cos theta )` `:. sqrt(x^(2)_y^(2))|_("min")=sqrt(365-28xx13)=sqrt(365-364)=1` |
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| 43. |
If a circle having centre at `(alpha,beta)` radius r completely lies with in two lines x+y=2 and x+y=-2, then, min. `(|alpha+beta+2|,|alpha+beta-2|` isA. greater than `sqrt2r`B. less than `sqrt2r`C. greater than 2rD. less than 2r |
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Answer» Correct Answer - or min. `{|alpha+beta+2|,|alpha+beta-2|}gtsqrt2r` |
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| 44. |
Let `xa n dy`be real variables satisfying `x^2+y^2+8x-10 y-40=0`. Let `a=max{sqrt((x+2)^2+(y-3)^2)}`and `b=min{sqrt((x+2)^2+(y-3)^2)}`. Then`a+b=18`(b) `a+b=sqrt(2)``a-b=4sqrt(2)`(d) `adotb=73`A. a+b=18B. `a-b=4sqrt2`C. `a+b=4sqrt2`D. a.b=73 |
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Answer» Correct Answer - `:.a+b=18,a-b=4sqrt2,ab=73` |
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| 45. |
If point `P(x,y)` is called a lattice point if `x, y in I.` Then the total number of lattice points in the interior of the circle `x^2 + y^2 =a^2, a != 0` can not be:A. 202B. 203C. 204D. 205 |
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Answer» Correct Answer - `:.` Number of such points must be of the form 4n+1, where n=0,1,2,….. |
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| 46. |
A region in the `x-y`plane is bounded by the curve `y=sqrt(25-x^2)`and the line `y=0`. If the point `(a ,a+1)`lies in the interior of the region, then`a in (-4,3)`(b) `a in (-oo,-1) in (3,oo)``a in (-1,3)`(d) none of theseA. `ain(-4,3)`B. `ain(-oo,-1)uu(3,oo)`C. `ain(-1,3)`D. None of these |
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Answer» Correct Answer - C |
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| 47. |
Show that the circle x2 + y2 – x + 4y + 4 = 0 touches y-axis. |
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Answer» Here f= 2 & C = 4 & f2 = C ⇒ (2)2 = 4 ⇒ Circle touches y-axis |
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| 48. |
Show that the circle x2 + y2 – 4x + 4y + 4 = 0 touches x-axis. |
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Answer» Here g = -2, C = 4 Here (-2)2 = 4 ⇒ g2 = C ∴ The circle touches x-axis. |
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| 49. |
Find the equation of the circle whose centre is same as the centre of the circle x2 + y2 + 6x + 2y + 1 = 0, and passing through the point (-2, -3). |
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Answer» Given x2 + y2 + 6x + 2y + 1 = 0, P = (-2,-3) Centre = C(-3,-1), Let the equations of the required circle is x2 + y2 + 6x + 2y + C = 0 r = CP = \(\sqrt{(-2 + 3)^2 + (-3 + 1)^2} = \sqrt{1^2 + 0} = 1\) ∴ r = \(\sqrt{g^2 + f^2 - c}\) I = \(\sqrt{9 + 1 - c}\) ⇒ 1 = 10 - c ⇒ c = 9 ∴ The required equation of the circle is x2 + y2 + 6x + 2y + 9 = 0. |
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| 50. |
A possible equation of L is (A) x 3y 1 (B) x 3y 1 (C) x 3y 1 (D) x 3y 5A. `x-sqrt(3) y =1`B. `x+ sqrt(3) y =1`C. `x-sqrt(3) y = -1`D. `x+sqrt(3) y =5` |
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Answer» Correct Answer - 1 The equation of tangent at `P(sqrt(3),1)` is `sqrt(3)x+y=4` The slope of line perpendicular to the above tangent is `1//sqrt(3)`. So, the equations of tangents with slope `1//sqrt(3)` to `(x-3)^(2)+y^(2)=1` will be `y= (1)/(sqrt(3))(x-3)=- 1sqrt(1+(1)/(3))` or `sqrt(3) y = x-3+- (2)` i.e., `sqrt(3) y = x-1 ` or `sqrt(3) y = x-5` |
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