Circle And System Of Circles Question 184
Question: The equation of the circumcircle of the triangle formed by the lines $ x=0,y=0,2x+3y=5 $ is
[MP PET 2004]
Options:
A) $ x^{2}+y^{2}+2x+3y-5=0 $
B) $ 6(x^{2}+y^{2})-5(3x+2y)=0 $
C) $ x^{2}+y^{2}-2x-3y+5=0 $
D) $ 6(x^{2}+y^{2})+5(3x+2y)=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
Given, triangle formed by the lines $ x=0 $ , $ y=0 $ , $ 2x+3y=5 $ , so vertices of the triangle are (0, 0), (5/2, 0) and (0, 5/3). Since circle is passing through (0, 0).
$ \therefore $ Equation of circle will be $ x^{2}+y^{2}+2gx+2fy=0 $ ..(i)
Also, circle is passing through (5/2, 0) and (0, 5/3) So, $ g=-5/4 $ , $ f=-5/6 $ .
Put the values of g and f in equation (i). After
Solving, we get $ 6(x^{2}+y^{2})-5(3x+2y)=0 $ , which is the required equation of the circle.
BETA
