The equation of the incircle of the triangle | vedclass

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Question

(C) The lines $x=0$ (y-axis) and $y=0$ (x-axis) form a right-angled triangle with the line $3x+4y-24=0$.Let the radius of the incircle be $r$. Since t

Vedclass · Accepted Answer

$x^2+y^2-4x-4y+4=0$

Vedclass · Answer

(C) The lines $x=0$ (y-axis) and $y=0$ (x-axis) form a right-angled triangle with the line $3x+4y-24=0$.Let the radius of the incircle be $r$. Since the circle is in the first quadrant and touches both axes,its center is $(r, r)$.The distance from the center $(r, r)$ to the line $3x+4y-24=0$ must be equal to the radius $r$.Using the perpendicular distance formula: $\left|\frac{3r+4r-24}{\sqrt{3^2+4^2}}\right| = r$.$\left|\frac{7r-24}{5}\right| = r$.Case $1$: $7r-24 = 5r$ $\Rightarrow 2r = 24$ $\Rightarrow r = 12$.Case $2$: $7r-24 = -5r$ $\Rightarrow 12r = 24$ $\Rightarrow r = 2$.Since the incircle must lie within the triangle,the radius $r=12$ is rejected as it lies outside the triangle. Thus,$r=2$.The equation of the circle is $(x-2)^2 + (y-2)^2 = 2^2$.$x^2 - 4x + 4 + y^2 - 4y + 4 = 4$.$x^2 + y^2 - 4x - 4y + 4 = 0$.Therefore,the correct option is $C$.

The equation of the incircle of the triangle formed by the lines $x=0, y=0$ and $3x+4y-24=0$ is

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