Question

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

• A. \(25\)
• B. \(10\)
• C. \(20\)
• D. \(400\)

Hint:

For a right-angled triangle, the circumcenter is the midpoint of the hypotenuse and the circumradius is half the length of the hypotenuse.

Answer 1

The Correct Option is B

Step 1: Find the intercepts of the line with the axes.
The equation of the line is 2x + 5y + a = 0.
Since the intercepts are with the positive coordinate axes, we must have a<0.
For x-intercept, put y = 0:
2x + a = 0
x = -\(\frac{a}{2}\)
So, the x-intercept is A(-\(\frac{a}{2}\), 0).
For y-intercept, put x = 0:
5y + a = 0
y = -\(\frac{a}{5}\)
So, the y-intercept is B(0, -\(\frac{a}{5}\)).

Step 2: Recognize the triangle formed.
The triangle formed by the line and the positive coordinate axes is a right-angled triangle with vertices A(-\(\frac{a}{2}\), 0), B(0, -\(\frac{a}{5}\)), and O(0, 0).

Step 3: Find the circumcenter and circumradius.
For a right-angled triangle, the circumcenter is the midpoint of the hypotenuse AB.
Circumcenter = \(\left(\frac{-\frac{a}{2} + 0}{2}, \frac{0 - \frac{a}{5}}{2}\right) = \left(-\frac{a}{4}, -\frac{a}{10}\right)\)
Circumradius (R) is half the length of the hypotenuse AB.
AB = \(\sqrt{\left(-\frac{a}{2} - 0\right)^2 + \left(0 - (-\frac{a}{5})\right)^2} = \sqrt{\frac{a^2}{4} + \frac{a^2}{25}} = \sqrt{\frac{25a^2 + 4a^2}{100}} = \sqrt{\frac{29a^2}{100}} = \frac{|a|\sqrt{29}}{10}\)
Circumradius (R) = \(\frac{AB}{2} = \frac{|a|\sqrt{29}}{20}\)

Step 4: Use the given area of the circum-circle.
Area of the circum-circle = \(\pi R^2 = \frac{29\pi}{4}\)
\(\pi \left(\frac{|a|\sqrt{29}}{20}\right)^2 = \frac{29\pi}{4}\)
\(\frac{a^2 \times 29}{400} = \frac{29}{4}\)
\(a^2 = \frac{29}{4} \times \frac{400}{29} = 100\)
|a| = \(\sqrt{100} = 10\)
Therefore, |a| = 10.