Circle And System Of Circles Question 100
Question: The lines $ 2x-3y=5 $ and $ 3x-4y=7 $ are the diameters of a circle of area 154 square units. The equation of the circle is
[IIT 1989; AIEEE 2003; Kerala (Engg.) 2005]
Options:
A) $ x^{2}+y^{2}+2x-2y=62 $
B) $ x^{2}+y^{2}-2x+2y=47 $
C) $ x^{2}+y^{2}+2x-2y=47 $
D) $ x^{2}+y^{2}-2x+2y=62 $
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Answer:
Correct Answer: B
Solution:
Centre of circle = Point of intersection of diameters = (1, -1) Now area $ =154\Rightarrow \pi r^{2}=154\Rightarrow r=7 $
Hence the equation of required circle is $ {{(x-1)}^{2}}+{{(y+1)}^{2}}=7^{2}\Rightarrow x^{2}+y^{2}-2x+2y=47 $ .
BETA
