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Conic Sections Test - 4

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Conic Sections Test - 4
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Weekly Quiz Competition
  • Question 1
    1 / -0

    The line 3x – 4y = 0

    Solution

    x2+y2=25

    After completing the square,we get

    (x−0)2+(y−0)2=52

    so center is (0,0) and radius is 5 units.

    As the normal should pass through the centre of the circle, center should satisfy the equation of normal

    so 3x – 4y = 0 is the equation of normal as (0,0) satisfy this equation.

  • Question 2
    1 / -0

    The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is

    Solution

    x2+y2=4a2

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