If the area of the circum-circle of the trian | vedclass

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(B) The line $2x + 5y + \alpha = 0$ intersects the coordinate axes at points $A$ and $B$. Since the triangle is formed with the positive coordinate ax

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$10$

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(B) The line $2x + 5y + \alpha = 0$ intersects the coordinate axes at points $A$ and $B$. Since the triangle is formed with the positive coordinate axes,the intercepts must be positive. Let $\alpha = -k$ where $k > 0$. The equation becomes $2x + 5y = k$,or $\frac{x}{k/2} + \frac{y}{k/5} = 1$.Thus,the vertices of the right-angled triangle are $O(0, 0)$,$A(\frac{k}{2}, 0)$,and $B(0, \frac{k}{5})$.The hypotenuse $AB$ is the diameter of the circum-circle. The length of the hypotenuse is $d = \sqrt{(\frac{k}{2})^2 + (\frac{k}{5})^2} = \sqrt{\frac{k^2}{4} + \frac{k^2}{25}} = \sqrt{\frac{29k^2}{100}} = \frac{k\sqrt{29}}{10}$.The radius $r$ of the circum-circle is $\frac{d}{2} = \frac{k\sqrt{29}}{20}$.The area of the circum-circle is $\pi r^2 = \pi \left(\frac{k^2 \cdot 29}{400}\right) = \frac{29\pi k^2}{400}$.Given the area is $\frac{29\pi}{4}$,we have $\frac{29\pi k^2}{400} = \frac{29\pi}{4}$.This simplifies to $k^2 = 100$,so $k = 10$.Since $k = |\alpha|$,we have $|\alpha| = 10$.

If the area of the circum-circle of the triangle formed by the line $2x + 5y + \alpha = 0$ and the positive coordinate axes is $\frac{29\pi}{4}$ sq. units,then $|\alpha| =$

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