If the chord x+y=1 of the circle x^2+y^2=a^2 | vedclass

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(C) The equation of the circle is $x^2+y^2=a^2$ and the chord is $x+y=1$. To find the equation of the pair of lines joining the origin to the intersec

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(C) The equation of the circle is $x^2+y^2=a^2$ and the chord is $x+y=1$. To find the equation of the pair of lines joining the origin to the intersection points of the circle and the chord,we homogenize the circle equation using the chord equation: $x^2+y^2=a^2(1)^2$ $x^2+y^2=a^2(x+y)^2$ $x^2+y^2=a^2(x^2+y^2+2xy)$ $(1-a^2)x^2 - 2a^2xy + (1-a^2)y^2 = 0$. Since the chord subtends a right angle at the origin,the sum of the coefficients of $x^2$ and $y^2$ must be zero: $(1-a^2) + (1-a^2) = 0$ $2(1-a^2) = 0$ $1-a^2 = 0$ $a^2 = 1$ Since $a$ represents a radius,$a=1$.

If the chord $x+y=1$ of the circle $x^2+y^2=a^2$ subtends a right angle at the origin,then $a=$

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