The equation of a circle touching the coordinate axes and the line $3 x-4 y=12$ is

The equation of a circle touching the coordinate axes and the line $3 x-4 y=12$ is
  1. $x^2+y^2+6 x+6 y+9=0$
  2. $x^2+y^2+6 x+6 y-9=0$
  3. $x^2+y^2-6 x-6 y+9=0$
  4. $x^2+y^2-6 x-6 y-9=0$

Solution

Since, circle touches both coordinates axes, then centre will be $(h, h)$ and radius $=h$ $ \begin{aligned} & \therefore\left|\frac{3 h-4 h-12}{\sqrt{(3)^2+(-4)^2}}\right|=h \Rightarrow\left|\frac{-h-12}{5}\right|=h \\ & \Rightarrow \quad-h-12= \pm 5 h \\ & -12= \pm 5 h+h \\ & -12=6 h \text { or }-12=-4 h \\ & h=-2 \text { or } 3 \\ & h=3 \\ \end{aligned} $ $\therefore$ Equation of circle will be $ \begin{aligned} (x-3)^2+(y-3)^2 & =3^2 \\ \Rightarrow x^2-6 x+9+y^2-6 y+9 & =9 \\ \Rightarrow \quad x^2+y^2-6 x-6 y+9 & =0 \end{aligned} $

Asked in: AP EAMCET 2018 (22 Apr Shift 1)

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