Question

The equation of the circle which touches the lines $x=0,y=0$ and $4x+3y=12$ is

• A. $x^2+y^2-2x-2y-1=0$
• B. $x^2+y^2-2x-2y+3=0$
• C. $x^2+y^2-2x-2y+2=0$
• D. $x^2+y^2-2x-2y+1=0$
• E. $x^2+y^2-2x-2y-3=0$

Answer 1

The correct option is D $x^2+y^2-2x-2y+1=0$

Let the radius of the circle be $r$.

Since, circle touches the lines $x=0,y=0$, i.e.,

$x$-axis and $y$-axis, then its centre is $C\equiv (r,r)$

Now, $\text{radius}\left(PC\right)=$ Perpendicular distance from $\left(C\right)$ to the line

$4x+3y=12$

i.e., $r=\frac{|4r+3r-12|}{\sqrt{16+9}}$

$\Rightarrow 7r-12=\pm 5r$

So, $r=1$ or $6$

But $r\ne 6$

Required equation of circle is

$(x-1{)}^2+(y-1{)}^2=1$

$\Rightarrow x^2+1-2x+y^2+1-2y=1$

$\Rightarrow x^2+y^2-2x-2y+1=0$