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 Multiple Choice QuestionsMultiple Choice Questions

241.

The equation of a circle passing through origin and radius is a, is

  • (x - a)2 + (y - a)2 = a2

  • x2 + y2 = a2

  • (x - a)2 + y2 = a2

  • None of the above


242.

Equation of tangent to the circle x2 + y2 - 2x - 2y + 1 = 0 perpendicular to y = x is given by

  • x + y ± 1 = 0

  • x + y = 2 ± 3

  • x + y ± 3 = 0

  • None of these


243.

The locus of centre of circles which cuts orthogonally the circle x2 + y2 - 4x + 8 = 0 and touches x + 1 = 0, is

  • y2 + 6x + 7 = 0

  • x2 + y2 + 2x + 3 = 0

  • x2 + 3y + 4 = 0

  • None of the above


244.

The condition for the line lx + my + n = 0 to be a normal to x225 + y29 = 1 is

  • l29 + m2l25 = n2256

  • 9m2 + 25l2 = 256n2

  • l29 - m2l25 = n2256

  • None of these


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245.

The radical centre of the system of circles

            x2 + y2 + 4x + 7 = 0,

2(x2 + y2) + 3x + 5y + 9 = 0

and               x2 + y2 + y = 0 is

  • (- 2, - 1)

  • (1, - 2)

  • (- 1, - 2)

  • None of these


246.

The point on the straight line y = 2x + 11 which is nearest to the circle 16(x2 + y) + 32x - 8y - 50 = 0, is

  • 92, 2

  • 92, - 2

  • - 92, 2

  • - 92, - 2


247.

The locus of the extrimities of the latusrectum of the family of ellipses b2x2 + y2 = a2b2 having a given major axis, is

  • x2 ± ay = a2

  • y2 ± bx = a2

  • x2 ± by = a2

  • y2 ± ax = b2


248.

The number of common tangents to two circles x2 + y2 = 4 and x2 + y2 - 8x + 12 = 0 is

  • 1

  • 2

  • 3

  • 4


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249.

If the tangent at the point 2secθ, 3tanθ of the hyperbola x24 - y29 = 1 is parallel to 3x - y + 4 = 0, then the value of θ is

  • π4

  • π3

  • π6

  • π2


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250.

If an equilateral triangle is inscribed in the circle x2 + y2 = a2, the lenth of its each side is

  • 2a

  • 3a

  • 32a

  • 13a


B.

3a

Given equation of the circle is         x2 + y2 = a2where, radius = a

let ABC be the inscribed equnateral triangle and length of its each sides be l,

Area of ABC = 34l2Also, OD = a2 - l24     from ODB Area of OBC = 12 × l × a2 - l24Now, area of ABC = 3 × Area of OBC 34l2 = 3 × l2a2 - l24   3l = 34a2 - l2     3l2 = 94a2 - l2       l2 = 12a2 - 3l2     4l2 = 12a2       l2 = 3a2        l = 3a


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