Circle And System Of Circles Question 371
Question: If the centre of a circle is (-6, 8) and it passes through the origin, then equation to its tangent at the origin, is
[MNR 1976]
Options:
A) $ 2y=x $
B) $ 4y=3x $
C) $ 3y=4x $
D) $ 3x+4y=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
Centre (-6, 8), radius $ =\sqrt{6^{2}+8^{2}}=10 $ Equation of circle $ x^{2}+y^{2}+12x-16y=0 $ Equation of tangent at (0, 0) is $ 3x-4y=0 $ .
BETA
