Suppose O(0,0) is the origin and the line L = | vedclass

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(A) The equation of the pair of lines joining the origin $O(0,0)$ to the points of intersection $A$ and $B$ is obtained by homogenizing the curve equa

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$\frac{1}{\sqrt{2}}$

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(A) The equation of the pair of lines joining the origin $O(0,0)$ to the points of intersection $A$ and $B$ is obtained by homogenizing the curve equation $x^2 + y^2 - 2x - 4y + 2 = 0$ using the line $x + y = \lambda$,i.e.,$\frac{x+y}{\lambda} = 1$.Substituting this into the curve equation:$x^2 + y^2 - 2(x + 2y)(\frac{x+y}{\lambda}) + 2(\frac{x+y}{\lambda})^2 = 0$$\lambda^2(x^2 + y^2) - 2\lambda(x^2 + 3xy + 2y^2) + 2(x^2 + 2xy + y^2) = 0$$(\lambda^2 - 2\lambda + 2)x^2 + (-6\lambda + 4)xy + (\lambda^2 - 4\lambda + 2)y^2 = 0$Since $\angle AOB = 90^{\circ}$,the sum of the coefficients of $x^2$ and $y^2$ is zero:$(\lambda^2 - 2\lambda + 2) + (\lambda^2 - 4\lambda + 2) = 0$$2\lambda^2 - 6\lambda + 4 = 0 \Rightarrow \lambda^2 - 3\lambda + 2 = 0$$(\lambda - 1)(\lambda - 2) = 0 \Rightarrow \lambda = 1, 2$.The lines are $x + y - 1 = 0$ and $x + y - 2 = 0$.The distance between these parallel lines is $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} = \frac{|-1 - (-2)|}{\sqrt{1^2 + 1^2}} = \frac{1}{\sqrt{2}}$.

Suppose $O(0,0)$ is the origin and the line $L = x + y - \lambda = 0$ meets the curve $x^2 + y^2 - 2x - 4y + 2 = 0$ at $A$ and $B$. If $\angle AOB = 90^{\circ}$,then the distance between such lines $L = 0$ is

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