Question

The equation of the incircle formed by the coordinate axes and the line 4x + 3y = 6 is
(a) x² + y² − 6x −6y + 9 = 0
(b) 4 (x² + y² − x − y) + 1 = 0
(c) 4 (x² + y² + x + y) + 1 = 0
(d) none of these

Answer 1

(b) 4 (x² + y² − x − y) + 1 = 0

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



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

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:
$$\begin{gathered} \left(\frac{0+0+3}{6},\frac{0+3+0}{6}\right) \\ =\left(\frac{1}{2},\frac{1}{2}\right) \end{gathered}$$

The equation of the incircle:

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

Also, radius of the incircle = $\frac{\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}{s}$
Here, $s=\frac{a+b+c}{2}=\frac{{\displaystyle \frac{5}{2}}+{\displaystyle \frac{3}{2}}+2}{2}=3$

∴ Radius of the incircle = $\frac{\sqrt{3\left(3-a\right)\left(3-b\right)\left(3-c\right)}}{3}$

$$\begin{gathered} =\frac{\sqrt{3\left(3-{\displaystyle \frac{5}{2}}\right)\left(3-{\displaystyle \frac{3}{2}}\right)\left(3-2\right)}}{3} \\ =\frac{\sqrt{3\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{3}{2}}\right)}}{3} \\ =\frac{1}{2} \end{gathered}$$

The equation of circle:
${\left(x-\frac{1}{2}\right)}^2+{\left(y-\frac{1}{2}\right)}^2=\frac{1}{4}$
$\Rightarrow 4\left(x^2+y^2-x-y\right)+1=0$