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

As the circle lies in third quadrant, then the centre is (-a, -a).

Perpendicular Distance (Between a point and line) = , whereas the point is and the line is expressed as ax + by + c = 0

The line which touches the circle is 3x−4y+8=0, which is a tangent to the circle.

∴ The perpendicular distance = a units (radius of the circle)

Co-ordinates of the centre of the circle = (-2,-2)

Since, the equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)² + (y – k)² = r²

(x – (-2))² + (y – (-2))² = 2²

(x + 2)² + (y + 2)² = 4

x² + 4x + 4 + y² + 4y + 4 - 4 = 0

x² + y² + 4x + 4y + 4 = 0

The equation of the given circle is x² + y² + 4x + 4y + 4 = 0.