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Find the equation of the circle which to...

Find the equation of the circle which touches both the axes and the line `3x-4y+8=0` and lies in the third quadrant.

Text Solution

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Let ' a ' be the radius of the circle.

therefore centre of circle

`( -a,-a)`[because the circle touches both the axes and lies in the third quadrant]

Given that the line` 3 x-4 y+8=0 `touches the circle.

Therefore` 3 x-4 y+8=0 `is tangent to the circle.

As we know that the radius and a tangent to a circle are always perpendicular to each other.

therefore perpendicular distance from the centre of circle to the tangent = a units, i.e.,

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Knowledge Check

  • Find the equation of a circle which touches both the axes and the line 3 4 8 0 x y - + = and lies in the third quadrant.

    A
    `x^(2)+y^(2)+4x+4y-4=0`
    B
    `x^(2)+y^(2)-4x-4y+4=0`
    C
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    D
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    A
    `x^(2)+y^(2)+- ax+- ay+(a^(2))/(4)=0`
    B
    `x^(2)+y^(2)+ax+- ay +(a^(2))/(4)=0`
    C
    `x^(2)+y^(2)-ax +- ay + (a^(2))/(4)=0`
    D
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    A
    `x^(2)+y^(2)-x-y+1=0`
    B
    `x^(2)+y^(2)-2x-2y+1=0`
    C
    `x^(2)+y^(2)-12x-12y+36=0`
    D
    `x^(2)+y^(2)-6x-6y+36=0`

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