Question

Find the equation of the circle which touches both the axes and the line `3x-4y+8=0` 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. `x^(2)+y^(2)+2x+2y+4=0`
D. none of these

Answer 1

Correct Answer - B

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`