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Grade 12th passDiscuss with colleagues and IITians

find the eqn of the circle which passes through the points (3,-2) (-2,0) and has its centre on the line 2x-y=3

Profile image of maitreyee jagannath
11 Years agoGrade 12th pass
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1 Answer

Profile image of Arun
6 Years ago
Circle := (x - a)^2 + (y - b)^2 = r^2
Center C(a , b) = (a, 2a-3)
as 2a - b = 3. Ie,. b = 2a-3
Also given points P(3,2) and Q(-2,0) are on the circle.
Midpoint of PQ: R= (1/2, 1)
Slope of PQ = 2/5
So the Slope of CR = -5/2
Equation of CR :
(y -1)/(x-1/2) = -5/2
4y + 10x = 9
Intersection of CR & y=2x-3
is center = C(7/6, -2/3)
Radius=R.
R^2 = 4/9 + 361/36 = 377/366
.
So circle. (x- 7/6)^2 + (y + 2/3)^2 = 377/36
 
 

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