Question

The equation of the incircle fonned by the coordinate axes and the line $4x+3y=6$ is

• A. $x^2+y^2-6x-6y+9=0$
• B. $4(x^2+y^2-x-y)+1=0$
• C. $4(x^2+y^2+x+y)+1=0$
• D. none of these

Answer 1

The correct option is B

$4(x^2+y^2-x-y)+1=0$

The line 4x + 3y = 6 cuts the coordinate
axes at $\left(\frac{3}{2}\right)$ and (0, 2)

The coordinates of the incentre is
$(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c})$

Here, $a=\frac{5}{2},b=\frac{3}{2},c=2,x_1=0,y_1=0,$
$x_2=0,y_2=2,x_3=\frac{3}{2},y_3=0$

Thus, the coordinates of the incentre:

$(\frac{0+0+3}{6},\frac{0+3+0}{6})$

$=(\frac{1}{2},\frac{1}{2})$

The equation of the incircle:

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

Also radius of the incircle
$=\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s}$

Here, $s=\frac{a+b+c}{2}=\frac{\frac{5}{2}+\frac{3}{2}+2}{2}=3$

$\therefore$ Radius of the incircl

$=\frac{\sqrt{3(3-\frac{5}{2})(3-\frac{3}{2})(3-2)}}{3}$

$=\frac{\sqrt{3\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)}}{3}$

$=\frac{1}{2}$

The equation of circle:

$={(x-\frac{1}{2})}^2+{(y-\frac{1}{2})}^2=\frac{1}{4}$

$\Rightarrow 4(x^2+y^2-x-y)+1=0$