The equation of image of pair of lines $y = | x - 1 |$ with respect to y-axis is

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

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

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4 x^{2} - 4 x - y^{2} + 1 = 0

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

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Solution

The correct option is C $x^{2} - y^{2} + 2 x + 1 = 0$
We have $y = | x - 1 |$

$\Rightarrow y^{2} = (x - 1)^{2}$

Change $x$ by $- x$ , then the required image is

$y^{2} = (- x - 1)^{2}$

$\Rightarrow y^{2} = x^{2} + 2 x + 1$

$\Rightarrow x^{2} - y^{2} + 2 x + 1 = 0$

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Find the equation of circle which touches $2 x - y + 3 = 0$ and pass through the points of intersection of the line $x + 2 y - 1 = 0$ and the circle $x^{2} + y^{2} - 2 x + 1 = 0$

Q. I : The equations to the direct common tangents to the circles

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

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y - 1 = 0

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Q. The circles

x^{2} + y^{2} - 2 x - 4 y + 1 = 0

and

x^{2} + y^{2} + 4 x + 4 y - 1 = 0

Find the equation of the circle passing the point (1, 2) and touching the line 2x + y - 1 = 0 at the point (1, -1).

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The equation of image of pair of lines y=|x−1| with respect to y-axis is

The correct option is C x2−y2+2x+1=0We have y=|x−1|⇒y2=(x−1)2Change x by −x, then the required image is y2=(−x−1)2⇒y2=x2+2x+1⇒x2−y2+2x+1=0

The equation of image of pair of lines $y = | x - 1 |$ with respect to y-axis is

The correct option is C $x^{2} - y^{2} + 2 x + 1 = 0$
We have $y = | x - 1 |$

$\Rightarrow y^{2} = (x - 1)^{2}$

Change $x$ by $- x$ , then the required image is

$y^{2} = (- x - 1)^{2}$

$\Rightarrow y^{2} = x^{2} + 2 x + 1$

$\Rightarrow x^{2} - y^{2} + 2 x + 1 = 0$