1sir isn't it that line is not x+y=2...so?why interchagne x and y!
Find the eqn of the circle on the other side of the line x+y=2 similarly situated as the circle x2+y2-2x=0.
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4 Answers
62you mean the mirror image?
then you have to interchange x and y
Or otherwise you will to find the mirror image of the center of this circle and then the radius is the same...
Answer will be x^2+y^2-2y=0
30SInce the line x+y-2=0 acts as the mirror, line joining the two centres of the circle is bisected perpendicularly by the mirror line x+y-2=0.
Centre of (x2+y2-2x=0) ≡ (1,0)
The line joining the centres intersects the mirror at (3/2,1/2)
Let the centre of the required circle be (x,y).
(1+x)/2=3/2 => x = 2
(0+y)/2=1/2 => y= 1
clearly,
equation of reqd. circle ≡ (x - 2)2 + (y - 1)2 = 1