CircleMedium
Question
The equation of the circle having the lines y2 - 2y + 4x - 2xy = 0 as its normals & passing through the point (2, 1) is -
Options
A.x2 + y2 - 2x - 4y + 3 = 0
B.x2 + y2 - 2x + 4y - 5 = 0
C.x2 + y2 + 2x + 4y - 13 = 0
D.none
Solution
y2 - 2xy + 4x - 2y = 0
y(y - 2x) - 2(y - 2x) = 0
⇒ y = 2 and y = 2x are the normals.
Now point of intersection of normals will give the centre of the circle i. e. (1, 2)
Radius of circle will be √2
∴ equation of circle : (x - 1)2 + (y - 2)2 = 2
y(y - 2x) - 2(y - 2x) = 0
⇒ y = 2 and y = 2x are the normals.
Now point of intersection of normals will give the centre of the circle i. e. (1, 2)
Radius of circle will be √2
∴ equation of circle : (x - 1)2 + (y - 2)2 = 2
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