Find the condition that the pair of lines joi | vedclass

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(A) The equation of the line is $y = mx + c$,which can be written as $\frac{y - mx}{c} = 1$. The equation of the circle is $x^{2} + y^{2} = a^{2}$. To

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

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(A) The equation of the line is $y = mx + c$,which can be written as $\frac{y - mx}{c} = 1$. The equation of the circle is $x^{2} + y^{2} = a^{2}$. To find the joint equation of the lines passing through the origin and the intersection points,we homogenize the circle equation using the line equation: $x^{2} + y^{2} = a^{2} \left( \frac{y - mx}{c} \right)^{2}$ $c^{2}(x^{2} + y^{2}) = a^{2}(y^{2} - 2mxy + m^{2}x^{2})$ $x^{2}(c^{2} - a^{2}m^{2}) + 2a^{2}mxy + y^{2}(c^{2} - a^{2}) = 0$. For these lines to be at right angles,the sum of the coefficients of $x^{2}$ and $y^{2}$ must be zero: $(c^{2} - a^{2}m^{2}) + (c^{2} - a^{2}) = 0$ $2c^{2} - a^{2}(1 + m^{2}) = 0$ $2c^{2} = a^{2}(1 + m^{2})$.

Find the condition that the pair of lines joining the origin to the points of intersection of the line $y = mx + c$ and the circle $x^{2} + y^{2} = a^{2}$ are at right angles to each other.

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