Circle And System Of Circles Question 369
Question: The points of intersection of the line $ 4x-3y-10=0 $ and the circle $ x^{2}+y^{2}-2x+4y-20=0 $ are
[IIT 1983]
Options:
A) $ (-2,-6),(4,2) $
B) $ (2,,6),(-4,-2) $
C) $ (-2,,6),(-4,,2) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Substituting $ x=\frac{3y+10}{4} $ in equation of circle, we get a quadratic in y.
Solving, we get two values of y as 2 and - 6 from which we get value of x.
BETA
