Conic-Sections Question 579
Question: An equilateral triangle is inscribed in the circle $ x^{2}+y^{2}=a^{2} $ with one of the vertices at (a, 0). What is the equation of the side opposite to this vertex?
Options:
A) $ 2x-a=0 $
B) $ x+a=0 $
C) $ 2x+a=0 $
D) $ 3x-2a=0 $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Since the equilateral triangle is inscribed in the circle with centre at the origin, centroid lies on the origin. So, $ \frac{AO}{OD}=\frac{2}{1} $ and $ OD=\frac{1}{2}AO=\frac{a}{2} $ So, other vertices of triangle have coordinates, $ ( -\frac{a}{2},\frac{\sqrt{3a}}{2} ) $ and $ [ -\frac{a}{2},-\frac{\sqrt{3}}{2}a ] $ $ ( -\frac{a}{2}\frac{\sqrt{3a}}{2} )y $ $ ( -\frac{a}{2},\frac{-\sqrt{3a}}{2} ) $
$ \therefore $ Equation of line BC is: $ x=-\frac{a}{2} $
$ \Rightarrow ,2x+a=0 $
BETA
