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

Options

A. `x^(2)+y^(2)-4x-4y+4=0`
B. `x^(2)+y^(2)+4x+4y+4=0`
C. `x^(2)+y^(2)+2x+2y+4=0`
D. `"none of these"`

Answer

B

Explanation

Let a be the radius of the circle. Then, the coordinates of the circle are (-a, -a). Now perpendicular distance from `C` to the line `AB` = Radius of the circle
`d=abs((-3a+4a+8)/sqrt(9+16))=abs((a+8)/5)`
`because a=pm((a+8)/5)`
Taking positive sign, `a=(a+8)/5`
`rArr` `5a=a+8`
`rArr` `4a=8` `rArr`a=2
Taking negative sign, `a=(-a-8)/5`
`rArr` `5a=-a-8 `
`rArr` `6a=-8` `rArr` `a=- 4/3`
But a`ne` `-4/3`
`because` a=2
So, the equation of the circle is
`(x+2)^(2)+(y+2)^(2)=2^(2)`
`rArr x^(2)+4x+4+y^(2)+4y+4=4`
`rArr x^(2)+y^(2)+4x+4y+4=0`