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(B) Let the centre of the circle be $(h, k)$ and radius be $r$.Since the circle touches the lines $x = 0$ and $y = 0$,the centre must be at $(r, r)$ o

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the centre of the circle be $(h, k)$ and radius be $r$.Since the circle touches the lines $x = 0$ and $y = 0$,the centre must be at $(r, r)$ or $(r, -r)$ or $(-r, r)$ or $(-r, -r)$.Assuming the circle is in the first quadrant,the centre is $(r, r)$ and the equation is $(x - r)^2 + (y - r)^2 = r^2$.This circle also touches the line $3x + 4y - 4 = 0$.The perpendicular distance from the centre $(r, r)$ to the line is equal to the radius $r$:$\frac{|3r + 4r - 4|}{\sqrt{3^2 + 4^2}} = r$$\frac{|7r - 4|}{5} = r$$|7r - 4| = 5r$Case $1$: $7r - 4 = 5r$ $\Rightarrow 2r = 4$ $\Rightarrow r = 2$.The equation is $(x - 2)^2 + (y - 2)^2 = 2^2$,which simplifies to $x^2 + y^2 - 4x - 4y + 4 = 0$.Case $2$: $7r - 4 = -5r$ $\Rightarrow 12r = 4$ $\Rightarrow r = 1/3$.Comparing with the given options,the correct equation is $x^2 - 4x + y^2 - 4y + 4 = 0$.

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

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