Question

The equation of the circle inscribed in the triangle formed by the straight line $4x+3y=6$ and both the coordinate axes is

• A. $5x^2+5y^2-5x-5y+11=0$
• B. $x^2+y^2-6x-6y+18=0$
• C. $4x^2+4y^2-4x-4y+1=0$
• D. $2x^2+2y^2-4x-4y+1=0$

Answer 1

The correct option is C $4x^2+4y^2-4x-4y+1=0$

As the circle is inscribed in the triangle formed by the straight line $4x+3y=6$ and both the coordinate axes
Therefore, the coordinates of the centre is
$C=(a,a)$ and radius $=a,a>0$

Distance from the centre to the line $4x+3y-6=0$ is equal to the radius, so
$\left|\frac{4a+3a-6}{\sqrt{4^2+3^2}}\right|=a\Rightarrow 7a-6=\pm 5a\Rightarrow a=\frac{1}{2},3$

As $(0,0)$ and $(3,3)$ lie on the opposite side of the line, so
$a=\frac{1}{2}$

Hence, the equation of the required circle is
${(x-\frac{1}{2})}^2+{(y-\frac{1}{2})}^2={\left(\frac{1}{2}\right)}^2\therefore 4x^2+4y^2-4x-4y+1=0$