Circle-And-System-Of-Circles Question 374
Question: The equation of normal to the circle $ 2x^{2}+2y^{2}-2x-5y+3=0 $ at (1, 1) is
[MP PET 2001]
Options:
A) $ x-2y=2 $
B) $ x+2y=\pm ,2\sqrt{3} $
C) $ x+2y=\pm ,2\sqrt{5} $
D) $ x-2y=\pm ,2\sqrt{5} $
Show Answer
Answer:
Correct Answer: C
Solution:
Trick: Only (b) and (c) lines are parallel to $ x+2y+3=0 $ Also the line is a tangent to $ x^{2}+y^{2}=4 $ Its distance from (0, 0) should be 2. Therefore c is the answer. Alternative method: Centre of $ x^{2}+y^{2}=4 $ is (0, 0). Tangents which are parallel to $ x+2y+3=0 $ is $ x+2y+\lambda =0 $ …..(i) Perpendicular distance from (0,0) to $ x+2y+\lambda =0 $ should be equal to radius of circle, (Clearly radius = 2).
$ \therefore $ $ \frac{0+2\times ,0+\lambda }{\sqrt{1^{2}+2^{2}}}=,\pm ,2 $
$ \Rightarrow $ $ \lambda ,=\pm ,2\sqrt{5} $ Put the value of $ \lambda , $ in (i), tangents of circle are $ x+2y=,\pm ,2\sqrt{5}. $
BETA
