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If the straight line 2x+3y=1 intersects ...

If the straight line 2x+3y=1 intersects the circle `x^2+y^2=4` at the points A and B then find the equation of the circle having AB as diameter.

Text Solution

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The correct Answer is:
`13(x^2+y^2)-4x-6y-50=0`
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Knowledge Check

  • If x - y + 1 = 0 meets the circle x^(2) + y^(2) + y - 1 = 0 at A and B , then the equation of the circle with AB as diameter is

    A
    `2(x^(2) + y^(2)) + 3x - y + 1 = 0 `
    B
    `2(x^(2) + y^(2)) + 3x -y + 2 = 0 `
    C
    `2(x^(2) + y^(2)) + 3x - y + 3 = 0 `
    D
    `x^(2) + y^(2) + 3x - y + 1 = 0 `

  • If the line x+y +1=0 intersects the circle x^2+y^2+x+3y=0 at two points A and B , then the centre of the circle which passes through the points A,B and the point of intersection of the tangents drawn at A and B to the given circle is

    A
    `(5/8,5/8)`
    B
    `(1,-1)`
    C
    `(3/4,-1/4)`
    D
    `(3,-4)`

  • If the circle x^2+y^2+2x+3y+1=0 cuts another circle x^2+y^2+4x+3y+2=0 in A and B , then the equation of the circle with AB as a diameter is

    A
    `2x^2+2y^2+2x+6y+1=0`
    B
    `x^2+y^2+x+3y+3=0`
    C
    `x^2+y^2+x+6y+1=0`
    D
    `2x^2+2y^2+x+3y+1=0`

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    If x+y=3 is the equation of the chord AB of the circle x^2+y^2-2x+4y-8=0 , find the equation of the circle having as diameter.

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