The image of every point lying on the curve x | vedclass

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(D) Let $P(x_0, y_0)$ be a point on the circle $x^2+y^2=1$. Let $P'(h, k)$ be its image in the line $x+y-1=0$.By the formula for the image of a point

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$x^2+y^2-2x-2y+1=0$

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(D) Let $P(x_0, y_0)$ be a point on the circle $x^2+y^2=1$. Let $P'(h, k)$ be its image in the line $x+y-1=0$.By the formula for the image of a point $(x_0, y_0)$ in the line $ax+by+c=0$:$\frac{h-x_0}{a} = \frac{k-y_0}{b} = -2 \frac{ax_0+by_0+c}{a^2+b^2}$Here,$a=1, b=1, c=-1$. So,$\frac{h-x_0}{1} = \frac{k-y_0}{1} = -2 \frac{x_0+y_0-1}{1^2+1^2} = -(x_0+y_0-1) = 1-x_0-y_0$From this,$h = x_0 + 1 - x_0 - y_0 = 1-y_0$ and $k = y_0 + 1 - x_0 - y_0 = 1-x_0$.Thus,$x_0 = 1-k$ and $y_0 = 1-h$.Since $(x_0, y_0)$ lies on $x^2+y^2=1$,we have:$(1-k)^2 + (1-h)^2 = 1$$1 - 2k + k^2 + 1 - 2h + h^2 = 1$$h^2 + k^2 - 2h - 2k + 1 = 0$Replacing $(h, k)$ with $(x, y)$,the locus is $x^2+y^2-2x-2y+1=0$.

The image of every point lying on the curve $x^2+y^2=1$ in the line $x+y=1$ satisfies the equation:

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