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

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

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

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

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

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Solution

The correct option is D $x^{2} - 4 x + y^{2} - 4 y + 4 = 0$
R.E.F image
Form figure ,radius of circle = h=R=CP
cp= $\frac{3 h + 4 R - 4}{5} = h = R$
$\Rightarrow 3 h + 4 R - 4 = 5 h$ and $3 h + 4 R - 4 = 5 R$
2h-4R+4=0 and 3h+R-4=0
$\Rightarrow$ h-2R+2=0 $\Rightarrow$ R=3h-4
h-6h+8+2=0
$\Rightarrow 5 h = 10 \Rightarrow h = 2, R = 2$
$\Rightarrow r a d i u s = 2, E q^{n} : (x - 2)^{2} + (y - 2)^{2} = 4$
$\Rightarrow x^{2} + y^{2} - 4 x - 4 y + 4 = 0 \Rightarrow (D)$

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

The correct option is D x2−4x+y2−4y+4=0R.E.F image Form figure ,radius of circle = h=R=CPcp=3h+4R−45=h=R⇒3h+4R−4=5h and 3h+4R−4=5R2h-4R+4=0 and 3h+R-4=0⇒ h-2R+2=0 ⇒R=3h-4 h-6h+8+2=0⇒5h=10⇒h=2,R=2⇒radius=2,Eqn:(x−2)2+(y−2)2=4⇒x2+y2−4x−4y+4=0⇒(D)

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