adas gupto

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.

4 Answers

62
Lokesh Verma ·

you 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

1
baburam ·

ans given is x2+y2-4x-2y+4=0

1
kunl ·

sir isn't it that line is not x+y=2...so?why interchagne x and y!

30
Ashish Kothari ·

SInce 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

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