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(B) Given curves are $x^2 + y^2 = a^2$ $(1)$ and $x^2 + y^2 - ax - ay = 0$ $(2)$.From $(1)$,we have $a^2 = x^2 + y^2$.Substituting this into $(2)$,we

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$xy = 0$

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(B) Given curves are $x^2 + y^2 = a^2$ $(1)$ and $x^2 + y^2 - ax - ay = 0$ $(2)$.From $(1)$,we have $a^2 = x^2 + y^2$.Substituting this into $(2)$,we get $x^2 + y^2 - ax - ay = 0$.Since $a^2 = x^2 + y^2$,we can write $a = \sqrt{x^2 + y^2}$.Alternatively,homogenizing the equation of the second curve using the first curve:$x^2 + y^2 - (ax + ay) \frac{x^2 + y^2}{a^2} = 0$.$a^2(x^2 + y^2) - (ax + ay)(x^2 + y^2) = 0$.$(x^2 + y^2)(a^2 - ax - ay) = 0$.This does not directly give the lines. Let us solve by intersection.Subtracting $(2)$ from $(1)$: $ax + ay = a^2 \implies x + y = a \implies a = x + y$.Substitute $a = x + y$ into $x^2 + y^2 = a^2$:$x^2 + y^2 = (x + y)^2 = x^2 + y^2 + 2xy$.$2xy = 0 \implies xy = 0$.Thus,the lines are $x = 0$ and $y = 0$.

The equation of the line joining the origin to the points of intersection of the curves $x^2 + y^2 = a^2$ and $x^2 + y^2 - ax - ay = 0$ is

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