Circle And System Of Circles Question 221
Question: The equation to the circle with centre (2, 1) and touching the line $ 3x+4y=5 $ is
[Karnataka CET 2005]
Options:
A) $ x^{2}+y^{2}-4x-2y+5=0 $
B) $ x^{2}+y^{2}-4x-2y-5=0 $
C) $ x^{2}+y^{2}-4x-2y+4=0 $
D) $ x^{2}+y^{2}-4x-2y-4=0 $
Show Answer
Answer:
Correct Answer: C
Solution:
Centre: (2, 1) Radius is distance of point (2, 1) from $ 3x+4y=5 $ i.e., $ | \frac{3.2+4.1-5}{\sqrt{9+16}} |\Rightarrow | \frac{6+4-5}{5} |=1 $ \ circle is $ {{(x-2)}^{2}}+{{(y-1)}^{2}}=1 $
$ x^{2}+y^{2}-4x-2y+4=0 $ .
BETA
