Circle And System Of Circles Question 162
Question: If the lines $ x+y=6 $ and $ x+2y=4 $ be diameters of the circle whose diameter is 20, then the equation of the circle is
Options:
A) $ x^{2}+y^{2}-16x+4y-32=0 $
B) $ x^{2}+y^{2}+16x+4y-32=0 $
C) $ x^{2}+y^{2}+16x+4y+32=0 $
D) $ x^{2}+y^{2}+16x-4y+32=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
Here $ r=10 $ (radius) Centre will be the point of intersection of the diameters, i.e. (8, -2).
Hence required equation is $ {{(x-8)}^{2}}+{{(y+2)}^{2}}=10^{2}\Rightarrow x^{2}+y^{2}-16x+4y-32=0 $ .
BETA
