Question

The equations of the circles, which touch both the axes and the line $4x+3y=12$ and have centres in the first quadrant, are

• A. $x^2+y^2+x-y+1=0$
• B. $x^2+y^2-2x-2y+1=0$
• C. $x^2+y^2-12x-12y+36=0$
• D. $x^2+y^2-6x-6y+36=0$

Answer 1

Answer: (B), (C)

$x^2+y^2-2x-2y+1=0$
$x^2+y^2-12x-12y+36=0$

Radius $\left(r\right)=$ perpendicular distance on line $4x+3y=12$ from centre
$\Rightarrow$ $r=r=\frac{|4r+3r-12|}{\sqrt{16+9}}$
$\Rightarrow |7r-12|=5r$
$\Rightarrow 7r-12=\pm 5r$
$\therefore 2r=12\Rightarrow r=6$
and $12r=12$

$\Rightarrow r=1$

(i) When centre is $(1,1)$ and radius is $1$, then equation of circle is
${(x-1)}^2+{(y-1)}^2=1$
$\Rightarrow x^2+y^2-2x-2y+1=0$

(ii) When centre is $(6,6)$ and radius is $2$, then equation of circle is
${(x-6)}^2+{(y-6)}^2=36$
$\Rightarrow x^2+y^2-12x-12y+36=0$