The point 3, - 4 lies on both&

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 Multiple Choice QuestionsMultiple Choice Questions

321.

If θ is the angle between the tangents from(- 1, 0) to the circle x2 + y2 - 5x + 4y - 2 = 0, then θ is equal to

  • 2tan-174

  • tan-174

  • 2cos-174

  • cot-174


322.

If 2x + 3y + 12 = 0 and x - y + 4λ = 0 are conjugate with respect to the parabola y = 8x, then λ is equal to

  • 2

  • - 2

  • 3

  • - 3


323.

For an ellipse with eccentricity 12 the centre is at the origin. If one directrix is x = 4, then the equation of the ellipse is

  • 3x2 + 4y2 = 1

  • 3x2 + 4y2 = 12

  • 4x2 + 3y2 = 1

  • 4x2 + 3y2 = 12


324.

The distance between the foci of the hyperbola x2 - 3y2 - 4x - 6y - 11 = 0 is

  • 4

  • 6

  • 8

  • 10


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325.

The radius of the circle with the polar equation

r2 - 8r(3cosθ + sinθ) + 15 = 0 is

  • 8

  • 7

  • 6

  • 5


326.

The area (in square unit) of the circle which touches the lines 4x + 3y = 15 and 4x + 3y = 5 is

  • 4π

  • π


327.

The equations of the circle which pass through the origin and makes intercepts of length 4 and 8 on the x and y-axes respectively are

  • x2 + y2 ± 4x ± 8y = 0

  • x2 + y2 ± 2x ± 4y = 0

  • x2 + y2 ± 8x ± 16y = 0

  • x2 + y2 ± x ± y = 0


328.

The locus of centre of a circle which passes through the origin and cuts off a length of 4 unit from the line x = 3 is

  • y2 + 6x = 0

  • y2 + 6x = 13

  • y2 + 6x = 10

  • x2 + 6y = 13


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329.

The diameters of a circle are along 2x +y - 7 and x + 3y - 11 = 0. Then, the equation of this circle, which also passes through (5, 7) is

  • x2 + y2 - 4x - 6y - 16 = 0

  • x2 + y2 - 4x - 6y - 20 = 0

  • x2 + y2 - 4x - 6y - 12 = 0

  • x2 + y2  + 4x  + 6y - 12 = 0


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330.

The point 3, - 4 lies on both the circlesx2 +y2 - 2x + 8y +13 = 0 and x2 +y2 -4x +6y + 11 = 0,Then, the angle between the circles is

  • 60°

  • tan-112

  • tan-135

  • 135°


D.

135°

Given circles are x2 + y2 - 2x +8y + 13 = 0and x2 + y2 - 4x +6y + 11 = 0Here, C1 = 1, - 4, C2 = 2, - 3,   r1 = 1 + 6 - 13 = 2and r2 = 4 + 9 - 11 = 2 cosθ = d2 - r12 - r222r1r2               = 2 - 4 - 22 × 2 × 2              = - 12      θ = 135°


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