Q89 / 160 · AP EAMCET 2024 (19 May Shift 2) · Mathematics
If the area of the circum-circle of triangle formed by the line $2 x+5 y+\alpha=0$ and the positive coordinate axes is $\frac{29 \pi}{4}$ sq.units, then $|\alpha|=$
- A$25$
- Correct answerB$10$
- C$20$
- D$400$
Solution
In the given figure $\angle A O B=90^{\circ}$
Since $2 x+5 y+a=0 \Rightarrow \frac{x}{\frac{-a}{2}}+\frac{y}{\frac{-a}{5}}=1$ So, $A\left(\frac{-a}{2}, 0\right), B\left(0, \frac{-a}{5}\right)$ Since, radius $=\frac{1}{2} \sqrt{\frac{a^2}{4}+\frac{a^2}{25}} \Rightarrow r=\frac{1}{2} \sqrt{\frac{29 a^2}{100}}=\frac{\sqrt{29}}{20} a$ Since, $\pi r^2=\frac{29 \pi}{4} \Rightarrow a^2=100 \Rightarrow|a|=10$.
Since $2 x+5 y+a=0 \Rightarrow \frac{x}{\frac{-a}{2}}+\frac{y}{\frac{-a}{5}}=1$ So, $A\left(\frac{-a}{2}, 0\right), B\left(0, \frac{-a}{5}\right)$ Since, radius $=\frac{1}{2} \sqrt{\frac{a^2}{4}+\frac{a^2}{25}} \Rightarrow r=\frac{1}{2} \sqrt{\frac{29 a^2}{100}}=\frac{\sqrt{29}}{20} a$ Since, $\pi r^2=\frac{29 \pi}{4} \Rightarrow a^2=100 \Rightarrow|a|=10$.
Hence, the correct answer is option: 'B'.