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The lines 2x -3y =5 and 3x-4y =7 are d...

The lines `2x -3y =5 ` and 3x-4y =7 are diameter of a circle having area as 154 sq. unit. Then , the equations of the circle is :

A

` x^(2)+ y^(2) +2x - 2y = 62`

B

` x ^(2) + y^(2)+ 2x- 2y= 47`

C

` x^(2) +y^(2)-2x +2y =47`

D

` x^(2)+ y^(2) - 2x + 2y =62`

Text Solution

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The correct Answer is:
C
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Knowledge Check

  • If the liens 3x - 4y - 7 = 0 and 2x - 3y - 5 = 0 are the two diameters of a circle of area 49 pi sq . units , the equation of the circle is

    A
    `x^(2) + y^(2) + 2x - 2y - 47 = 0 `
    B
    ` x^(2) + y^(2) + 2x - 2y - 62 = 0`
    C
    `x^(2) + y^(2) - 2x + 2y - 62 = 0 `
    D
    ` x^(2) + y^(2) - 2x + 2y - 47 = 0 `

  • The two diameters of a circle are segments of the straight lines x-y =5 and 2x+y =4 .If the radius of the circle is 5 then the equation of the circle is

    A
    `x^(2)+y^(2) -6x+4y =12 `
    B
    `x^(2) +y^(2) -3x+2y=12 `
    C
    `x^(2)+y^(2) -6x+2y=12 `
    D
    `x^(2)+y^(2) -8x +6y -18 =0`

  • If (3,-2) is the centre of a circle and 4x+3y +19 =0 is a tangent to the circle , then the equation of the circle is

    A
    `x^(2)+y^(2)-6x+4y+25=0`
    B
    `x^(2)+y^(2)-6x+4y+12=0`
    C
    `x^(2)+y^(2)-6x+4y-12=0`
    D
    `x^(2)+y^(2)-6x+4y+13=0`

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    In the figure, O is the centre of the circle and x^2+y^2=25 is the equation of the circle. Write the equation of the circle whose centre is at the origin and radius is 3. .

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