$C$ ircle touching both the axes and the line $3 x + 4 y = 12$ is

x^{2} + y^{2} + 2 x + 2 y + 1 = 0

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x^{2} + y^{2} - 2 x - 2 y + 1 = 0

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x^{2} + y^{2} - 4 x - 4 y + 4 = 0

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x^{2} + y^{2} + 6 x + 6 y + 9 = 0

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Solution

The correct option is B $x^{2} + y^{2} - 2 x - 2 y + 1 = 0$
Given circle touching both the axes, then centre of circle is at $(a, a)$ .
and circle also touches $3 x + 4 y = 12$
than diatance of $(a, a)$ from $3 s + 4 y = 12$ is equal to radius of circle.
radius of circle $= | a | = (\frac{3 a + 4 a - 12}{5})$
$| a | = ∣ ∣ ∣ \frac{7 a - 12}{5} ∣ ∣ ∣$
$5 a = 7 a - 12 o r - 5 a = 7 a - 12$
$a = + 1$ or $a = 1$
$(x - 1)^{2} + (y - 1)^{2} = 1$
$\Rightarrow x^{2} + y^{2} - 2 x - 2 y + 1 = 0$ .

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Circle touching both the axes and the line 3x+4y=12 is

$C$ ircle touching both the axes and the line $3 x + 4 y = 12$ is