JEE Mathematics Circles MCQs Set 21

Answer: (b) \(a^2\)

Explanation:
1. Orthogonality: \(2g_1g_2=-a^2-a^2\Rightarrow g_1g_2=-a^2\). Both circles pass through \((0,\pm a)\).
2. Check with \(a=1,\ g_1=1,\ g_2=-1\): centres \((\mp1,0)\), radius \(\sqrt2\). The common tangents are \(y=\pm\sqrt2\) and the perpendicular distances from \((0,1),(0,-1)\) to \(y=\sqrt2\) are \(\sqrt2-1\) and \(\sqrt2+1\).
3. Product \(=2-1=1=a^2\). In general \(p_1p_2=a^2\).

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Answer: (a) \(8\sqrt3\) sq. units

Explanation:
1. The circles are \((x\mp2)^2+y^2=16\), with centres \((\pm2,0)\). The common region meets the \(x\)-axis at \(x=\pm2\) (the diagonal of length \(4\)).
2. The other two vertices are where the circles meet: \(x=0,\ y^2=16-4=12\Rightarrow y=\pm2\sqrt3\), a diagonal of length \(4\sqrt3\).
3. Area \(=\tfrac12\cdot4\cdot4\sqrt3=8\sqrt3\).

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Answer: (b) \(c^2-34c+64=0\)

Explanation:
1. Centre \((-3,5)\), \(r^2=34-c\). The power of the origin is \(c\), so each tangent has length \(\sqrt c\).
2. The quadrilateral consists of two right triangles with legs \(r\) and \(\sqrt c\): area \(=r\sqrt c=8\).
3. \(r^2c=64\Rightarrow(34-c)c=64\Rightarrow c^2-34c+64=0\).

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Answer: (b) Circle

Explanation:
1. Let the chord end points be \(A,B\) (fixed for a given chord) and \(P\) move on the circle. The centroid is \(G=\dfrac{A+B+P}3\).
2. As \(P\) moves on the circle of radius \(5\), \(G\) moves on a circle of radius \(\dfrac53\) with centre \(\dfrac{A+B}3\).

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Answer: (b) \(x^2+y^2-x-y=0\)

Explanation:
1. The smallest circle through two points has them as the ends of a diameter.
2. \((x-1)(x-0)+(y-0)(y-1)=0\Rightarrow x^2+y^2-x-y=0\).

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Answer: (b) \(a=b\pm\sqrt2c\)

Explanation:
1. The centres are \((a,b)\) and \((b,a)\), at distance \(\sqrt2|a-b|\); both radii are \(c\).
2. Equal circles can touch only externally: \(\sqrt2|a-b|=2c\Rightarrow|a-b|=\sqrt2c\), so \(a=b\pm\sqrt2c\).

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Answer: (c) 16

Explanation:
1. \(n\alpha=8\) and \(\binom n2\alpha^2=24\Rightarrow\dfrac{8(8-\alpha)}2=24\Rightarrow\alpha=2,\ n=4\).
2. \(P=(2,4)\). By the power of a point, \(PA\cdot PB=2^2+4^2-4=16\).

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Answer: (c) 32 sq. unit

Explanation:
1. The centre lies on the perpendicular bisector of \(AB\): \(x=1\), and on \(4y=x+7\): \(y=2\). Centre \((1,2)\).
2. \(r^2=(1+3)^2+(2-4)^2=20\), so the diagonal is \(2\sqrt{20}\) and \(AB=8\).
3. \(BC=\sqrt{80-64}=4\), area \(=8\times4=32\).

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Answer: (a) \(a^{-2}+b^{-2}=c^{-1}\)

Explanation:
1. Centres \((-a,0),(0,-b)\), radii \(\sqrt{a^2-c},\sqrt{b^2-c}\), distance between centres \(\sqrt{a^2+b^2}\).
2. External touching: \(a^2+b^2=(a^2-c)+(b^2-c)+2\sqrt{(a^2-c)(b^2-c)}\Rightarrow c=\sqrt{(a^2-c)(b^2-c)}\).
3. Squaring: \(c^2=a^2b^2-c(a^2+b^2)+c^2\Rightarrow c(a^2+b^2)=a^2b^2\), i.e. \(\dfrac1{a^2}+\dfrac1{b^2}=\dfrac1c\).

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Answer: (a) \((-\infty,-\sqrt3]\cup[\sqrt3,+\infty)\)

Explanation:
1. The distance of the centre from \(mx-y+2=0\) is \(\dfrac2{\sqrt{1+m^2}}\), which must be \(\le1\).
2. \(4\le1+m^2\Rightarrow m^2\ge3\), so \(m\le-\sqrt3\) or \(m\ge\sqrt3\).

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Answer: (a) 1 : 2

Explanation:
1. On the given circle \(x^2+y^2=\dfrac{16}5x-\dfrac{64}{15}y\).
2. \(t_1^2=x^2+y^2-\tfrac{24}5x+\tfrac{32}5y+15=-\tfrac85x+\tfrac{32}{15}y+15\) and \(t_2^2=x^2+y^2-\tfrac{48}5x+\tfrac{64}5y+60=-\tfrac{32}5x+\tfrac{128}{15}y+60\).
3. Hence \(t_2^2=4t_1^2\), so \(t_1:t_2=1:2\).

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Answer: (a) \(x^2+y^2-2x-4y+1=0\)

Explanation:
1. The centre is the radical centre. Radical axes: \(S_1-S_2:\ x+y=3\); \(S_2-S_3:\ 3x-2y+1=0\). They meet at \((1,2)\).
2. The orthogonal circle has radius\(^2\) equal to the power of this point: \(S_1(1,2)=1+4-3-12+14=4\).
3. Circle: \((x-1)^2+(y-2)^2=4\Rightarrow x^2+y^2-2x-4y+1=0\).

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Answer: (c) \(\dfrac\pi4\)

Explanation:
1. \(\vec{OQ}\cdot\vec{OR}=-12+12=0\), so \(\angle QOR=90^\circ\).
2. The angle at a point of the major arc is half the angle at the centre: \(\angle QPR=45^\circ=\dfrac\pi4\).

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Answer: (a) \(h=\sqrt5\)

Explanation:
1. The origin lies on the \(y\)-axis tangent and on the secant \(AB\) (the \(x\)-axis), so the power of the origin gives \(OC^2=OA\cdot OB=1\cdot5\).
2. Hence \(h=\sqrt5\) (taking \(h>0\)); the circle touching the axis gives the largest angle subtended by \(AB\).

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Answer: (c) 3

Explanation:
1. The given circle has centre \((1,3)\) and radius \(2\). A diameter is a chord of the other circle, so the midpoint of that chord is \((1,3)\).
2. The distance between the centres is \(\sqrt{1+4}=\sqrt5\), which is perpendicular to the chord. Then \(R^2=(\sqrt5)^2+2^2=9\), \(R=3\).

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Answer: (a) 1

Explanation:
1. The locus of \(P\) with \(PA/PB=k\) is the Apollonius circle for \(k\ne1\).
2. For \(k=1\) the locus is the perpendicular bisector of \(AB\), a straight line, not a circle. So \(k\ne1\).

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Answer: (c) \(\dfrac{10}9\sqrt5\) sq. units

Explanation:
1. The chord of contact is \(3x=4\), i.e. \(x=\tfrac43\). Its half-length is \(\sqrt{4-\tfrac{16}9}=\dfrac{2\sqrt5}3\), so \(AB=\dfrac{4\sqrt5}3\).
2. The height from \(P\) is \(3-\tfrac43=\tfrac53\). Area \(=\tfrac12\cdot\dfrac{4\sqrt5}3\cdot\dfrac53=\dfrac{10\sqrt5}9\).

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Answer: (a) Externally at \((0,1)\)

Explanation:
1. Centres \((-1,1)\) and \((1,1)\), each radius \(1\). The distance between centres is \(2=1+1\): external contact.
2. The point of contact is the midpoint \((0,1)\).

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45. o
46. i
47. n
48. t
49.
50. o
51. f
52.
53. t
54. h
55. e
56.
57. c
58. e
59. n
60. t
61. r
62. e
63. s
64. .

Answer: (b) \(x^2+y^2=2\)

Explanation:
1. For a chord subtending \(90^\circ\) at the centre, the distance from the centre to the chord is \(\dfrac r{\sqrt2}=\sqrt2\).
2. The midpoint is at distance \(\sqrt2\) from the centre, so the locus is \(x^2+y^2=2\).

Teacher's Note:
1. T
2. h
3. e
4.
5. m
6. i
7. d
8. p
9. o
10. i
11. n
12. t
13.
14. o
15. f
16.
17. s
18. u
19. c
20. h
21.
22. a
23.
24. c
25. h
26. o
27. r
28. d
29.
30. i
31. s
32.
33. a
34. t
35.
36. d
37. i
38. s
39. t
40. a
41. n
42. c
43. e
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45. r
46. /
47. s
48. q
49. r
50. t
51. 2
52.
53. f
54. r
55. o
56. m
57.
58. t
59. h
60. e
61.
62. c
63. e
64. n
65. t
66. r
67. e
68. .

Answer: (b) \(ag+bf-c=0\)

Explanation:
1. A normal passes through the centre \((-g,-f)\).
2. \(-ag-bf+c=0\Rightarrow ag+bf-c=0\).

Teacher's Note:
1. N
2. o
3. r
4. m
5. a
6. l
7. s
8.
9. o
10. f
11.
12. a
13.
14. c
15. i
16. r
17. c
18. l
19. e
20.
21. p
22. a
23. s
24. s
25.
26. t
27. h
28. r
29. o
30. u
31. g
32. h
33.
34. i
35. t
36. s
37.
38. c
39. e
40. n
41. t
42. r
43. e
44. .

Answer: (b) \(2x-3y+10=0\)

Explanation:
1. The centre lies on \(y=2\): \((h,2)\) with \(r=|h|\). Through \((-2,4)\): \((h+2)^2+4=h^2\Rightarrow h=-2\). Centre \((-2,2)\).
2. A diameter passes through the centre: \(2(-2)-3(2)+10=0\), so \(2x-3y+10=0\) works; the other options do not vanish at \((-2,2)\).

Teacher's Note:
1. A
2.
3. d
4. i
5. a
6. m
7. e
8. t
9. e
10. r
11.
12. m
13. u
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15. t
16.
17. p
18. a
19. s
20. s
21.
22. t
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24. r
25. o
26. u
27. g
28. h
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30. t
31. h
32. e
33.
34. c
35. e
36. n
37. t
38. r
39. e
40. .