Conic Sections Question 488
Question: A square is inscribed in the circle $ x^{2}+y^{2}-2x+4y-93=0 $ with its sides parallel to the coordinate axes. The coordinates of its vertices are
Options:
A) (-6, -9), (-6, 5), (8, -9), (8, 5)
B) (-6, 9), (-6, -5), (8, -9), (8, 5)
C) (-6, -9), (-6, 5), (8, 9), (8, 5)
D) (-6, -9), (-6, 5), (8, -9), (8, -5)
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Let $ x=\alpha ,x=b,y=c $ and $ y=d $ be the sides of the square.
The length of each diagonal of the square is equal to the diameter of the circle, i.e., $ 2\sqrt{98.} $
Let $ l $ be the length of each side of the square. Then, $ 2l^{2}={{(Diagonal)}^{2}} $ or $ l=14 $
Therefore, each side of the square is at a distance 7 form the center (1,-2) of the given circle. This implies that $ a =-6,b=8, $
$ c=-9 $ , and d=5.
Hence, the vertices of the square are (-6,-9)(-6,5),(8,-9) and (8,5).
BETA
