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The radius of the circle passing through...

The radius of the circle passing through (6,2) and the equation of two normals for the circle are `x+y=6` and `x+2y=4` is

A

`sqrt(5)`

B

`2sqrt(5)`

C

`3sqrt(5)`

D

`4sqrt(5)`

Text Solution

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The correct Answer is:
B
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Knowledge Check

  • The equation of the circle passing through (1,2) and the points of intersection of the circles x^2+y^2-8x-6y+21=0 and x^2+y^2-2x-15=0 is

    A
    `x^2+y^2+6x-2y+9=0`
    B
    `x^2+y^2-6x-4y+9=0`
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    `x^2+y^2-6x-4y+9=0`
    D
    `x^2+y^2-6x+4y+9=0`

  • The equation of the circle passing through (0,0) and cutting orthogonally the circles x^(2) + y^(2) + 6x - 15 = 0 , x^(2) + y^(2) - 8y + 10 = 0 is

    A
    ` 2x^(2) + 2y^(2) - 10x - 5y = 0 `
    B
    `2x^(2) + 2y^(2) + 10x + 5y = 0 `
    C
    ` x^(2) + y^(2) - 5x + 5y = 0 `
    D
    `2x^(2) + 2y^(2) + 10x - 5y = 0 `

  • The equation of the circle passing through the point of intersection of the circles x^(2) + y^(2) + 6x + 4y - 12 = 0 , x^(2) + y^(2)- 4x - 6y - 12 = 0 and having radius sqrt(13) is

    A
    `9x^(2) + 9y^(2) + 16y - 34 = 0 `
    B
    `10(x^(2) + y^(2)) + 3y 0 86 = 0 `
    C
    `x^(2) +y^(2) - x- 2y = 0 `
    D
    `x^(2) +y^(2) - 2y - 12 = 0 `

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