If P1, P2, P3 are the perimeters of the three circles x2&nbs

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 Multiple Choice QuestionsMultiple Choice Questions

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291.

If P1, P2, P3 are the perimeters of the three circles 

x2 + y2 + 8x - 6y = 0, 4x2 + 4y - 4x - 12y - 186 = 0 and x2 + y - 6x + 6y - 9 = 0 respectively, then

  • P1 <  P2  <  P3

  • P1 <  P3  <  P2

  • P3 <  P2  <  P1

  • P2 <  P3  <  P1


B.

P1 <  P3  <  P2

Given circle are x2 + y2  + 8x - 6y = 04x2 + 4y2 - 4x - 12y - 186 = 0 and x2 + y2 - 6x + 6y - 9 = 0.Let r1,  r2 be the radius of the respective circle,thenr1 = - 42 + - 32 + 0  = 25 = 5r2 = 122 + 322 + 1864 = 49 = 7r3 = 32 + 32 + 9 = 27 = 33 P1 = 2πr1 = 10π      P2 = 2πr2 = 14π      P3 = 2πr3 = 63π P1 < P3  < P2


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292.

If the line 3x - 2y + 6 = 0 meets X-axis and Y-axis, respectively at A and B, then the equation of the circle with radius AB and centre at A is

  • x2 + y2 + 4x + 9 = 0

  • x2 + y2 + 4x - 9 = 0

  • x2 + y2 + 4x + 4 = 0

  • x2 + y2 + 4x - 4 = 0


293.

A line l meets the circle x2 + y2 = 61 in A, B and P(- 5, 6) is such that PA = PB = 10. Then,the equation of l is

  • 5x + 6y + 11 = 0

  • 5x - 6y - 11 = 0

  • 5x - 6y + 11 = 0

  • 5x - 6y + 11 = 0


294.

If (1, a), (b, 2) are conjugate points with respect to the circle x2 + y2 = 25, then 4a + 2b is equal to

  • 25

  • 50

  • 100

  • 150


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295.

The eccentricity of the conic 36x2 + 144y2 - 36x - 96y -119 = 0 is

  • 32

  • 12

  • 34

  • 13


296.

The polar equation cosθ + 7sinθ = 1r represents a

  • circle

  • parabola

  • straight line

  • hyperbola


297.

The centre of the circle r2 - 4rcosθ + sinθ - 4 = 0 in cartesian coordinates is

  • (1, 1)

  • (- 1, - 1)

  • (2, 2)

  • (- 2, - 2)


298.

The radius of the circle r = 3sinθ + cosθ is

  • 1

  • 2

  • 3

  • 4


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299.

The equation of the circle whose diameter is the common chord of the circles x2 + y2 + 2x + 3y + 2 = 0 and x2 + y2 + 2x - 3y - 4 = 0 is

  • x2 + y2 + 2x + 2y + 2 = 0

  • x2 + y2 + 2x + 2y - 1 = 0

  • x2 + y2 + 2x + 2y + 1 = 0

  • x2 + y2 + 2x + 2y + 3 = 0


300.

If x - y + 1 = 0 meets the circlex2 + y2 + y - 1 = 0 at A and B, then the equation of the circle with AB as diameter is

  • 2(x2 + y2) + 3x - y + 1 = 0

  • 2 (x2 + y2) + 3x - y + 2 = 0

  • 2(x2 + y2) + 3x - y + 3 = 0

  • x2 + y2 + 3x - y + 1 = 0


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