Circle And System Of Circles Question 354
The point of contact of the tangent to the circle $ x^{2}+y^{2}=5 $ at the point (1, -2) which touches the circle $ x^{2}+y^{2}-8x+6y+20=0 $ , is
[Roorkee 1989]
Options:
A) (2, -1)
B) (3, -1)
C) (4, -1)
D) (5, -1)
Show Answer
Answer:
Correct Answer: B
Solution:
Equation of the tangent at $ (1,\ -2) $ to the circle $ x^{2}+y^2=1 $ is $ x-2y=-5 $ . Now choose answers. Here, only point (3, -1) lies on the tangent.
BETA
