Circle-And-System-Of-Circles Question 375
Question: The square of the length of the tangent from (3, ?4) on the circle $ x^{2}+y^{2}-4x-6y+3=0 $ is
[MP PET 2000]
Options:
A) $ 2x+y=3 $
B) $ x-2y=3 $
C) $ x+2y=3 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
The equation of tangent to the given circle at $ (1,,1) $ is $ 2x+2y-(x+1)- $ $ \frac{5}{2},(y+1)+3=0 $
$ \Rightarrow $ $ x-\frac{1}{2},y-\frac{1}{2}=0 $
$ \Rightarrow $ $ 2x-y-1=0 $ Slope of tangent = 2,
$ \therefore $ Slope of normal $ =-\frac{1}{2} $ Hence equation of normal at (1, 1) is $ y-1=-\frac{1}{2}(x-1) $
$ \Rightarrow $ $ x+2y=3. $
BETA
