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(B) The given curve is $x^2 + y^2 - 2x - 1 = 0$ and the line is $x + y = 1$. To make the equation of the curve homogeneous of degree $2$,we rewrite th

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$	an^{-1}(2)$

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(B) The given curve is $x^2 + y^2 - 2x - 1 = 0$ and the line is $x + y = 1$. To make the equation of the curve homogeneous of degree $2$,we rewrite the line as $(x + y) = 1$. Substituting this into the curve equation: $x^2 + y^2 - 2x(x + y) - (x + y)^2 = 0$ $x^2 + y^2 - 2x^2 - 2xy - (x^2 + 2xy + y^2) = 0$ $-2x^2 - 4xy = 0$ $x^2 + 2xy = 0$ Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = 1, h = 1, b = 0$. The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$. $	an 	heta = \left| \frac{2\sqrt{1^2 - (1)(0)}}{1 + 0} ight| = 2$. Therefore,$	heta = 	an^{-1}(2)$.

The acute angle formed between the lines joining the origin to the points of intersection of the curves $x^2 + y^2 - 2x - 1 = 0$ and $x + y = 1$ is

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