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(C) To find the equation of the pair of lines joining the origin to the intersection points,we make the circle equation homogeneous using the line equ

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$2c^2 = a^2(1 + m^2)$

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(C) To find the equation of the pair of lines joining the origin to the intersection points,we make the circle equation homogeneous using the line equation $y - mx = c$,which implies $\frac{y - mx}{c} = 1$.Substituting this into the circle equation $x^2 + y^2 = a^2(1)^2$,we get:$x^2 + y^2 = a^2 \left( \frac{y - mx}{c} \right)^2$$c^2(x^2 + y^2) = a^2(y^2 - 2mxy + m^2x^2)$$(c^2 - a^2m^2)x^2 + 2a^2mxy + (c^2 - a^2)y^2 = 0$For these lines to be mutually perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero:$(c^2 - a^2m^2) + (c^2 - a^2) = 0$$2c^2 - a^2(m^2 + 1) = 0$$2c^2 = a^2(1 + m^2)$.

The lines joining the origin to the points of intersection of the line $y = mx + c$ and the circle $x^2 + y^2 = a^2$ will be mutually perpendicular,if

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