The condition that the lines joining the orig | vedclass

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(A) Given line: $\frac{x}{a} + \frac{y}{b} = 2 \Rightarrow \frac{x}{2a} + \frac{y}{2b} = 1$. Given circle: $(x - a)^2 + (y - b)^2 = r^2 \Rightarrow x^

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$a^2 + b^2 = r^2$

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(A) Given line: $\frac{x}{a} + \frac{y}{b} = 2 \Rightarrow \frac{x}{2a} + \frac{y}{2b} = 1$. Given circle: $(x - a)^2 + (y - b)^2 = r^2 \Rightarrow x^2 + y^2 - 2ax - 2by + a^2 + b^2 - r^2 = 0$. To find the pair of lines joining the origin to the intersection points,we homogenize the circle equation using the line equation: $x^2 + y^2 - 2(ax + by)(\frac{x}{2a} + \frac{y}{2b}) + (a^2 + b^2 - r^2)(\frac{x}{2a} + \frac{y}{2b})^2 = 0$. For the lines to be at right angles,the sum of the coefficients of $x^2$ and $y^2$ must be zero. Expanding the equation,the coefficient of $x^2$ is $1 - 1 + \frac{a^2 + b^2 - r^2}{4a^2} \cdot b^2$ (simplified logic leads to the condition): $a^2 + b^2 - r^2 = 0 \Rightarrow a^2 + b^2 = r^2$.

The condition that the lines joining the origin to the points of intersection of the line $\frac{x}{a} + \frac{y}{b} = 2$ and the circle $(x - a)^2 + (y - b)^2 = r^2$ are at right angles is

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