Q. 64.3( 7 Votes )

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

Answer :

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.


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