Question

The equation of a circle which touches both axes and the line $3 x - 4 y + 8 = 0$ and whose centre lies in the third quadrant is.

• A. $x ^ { 2 } + y ^ { 2 } - 4 x + 4 y - 4 = 0$
• B. $x ^ { 2 } + y ^ { 2 } - 4 x + 4 y + 4 = 0$
• C. $x ^ { 2 } + y ^ { 2 } + 4 x + 4 y + 4 = 0$
• D. $x ^ { 2 } + y ^ { 2 } - 4 x - 4 y - 4 = 0$

Answer 1

Answer: (C)

$x ^ { 2 } + y ^ { 2 } + 4 x + 4 y + 4 = 0$

Answer 2

The equation of circle in third quadrant touching the coordinate axes with centre $( - a , - a )$ and radius ‘a’ is $x ^ { 2 } + y ^ { 2 } + 2 a x + 2 a y + a ^ { 2 } = 0$ and we know

$\left| \frac { 3 ( - a ) - 4 ( - a ) + 8 } { \sqrt { 9 + 16 } } \right| = a \Rightarrow a = 2$

Hence the required equation is

$x ^ { 2 } + y ^ { 2 } + 4 x + 4 y + 4 = 0$.

Trick : Obviously the centre of the circle lies in III quadrant, which is given by (3).