Equation of circle having lines y^2 - 2y + 4x - 2xy = 0 as its normals & passing through point (

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10-1.Circle and System of Circles

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The equation of the circle having the lines $y^2 - 2y + 4x - 2xy = 0$ as its normals $\&$ passing through the point $(2 , 1)$ is :

A

$x^2 + y^2 - 2x - 4y + 3 = 0$

B

$x^2 + y^2 - 2x + 4y - 5 = 0$

C

$x^2 + y^2 + 2x + 4y - 13 = 0$

D

none

Solution

normals are $(y – 2)(y – 2x) = 0$ should pass through centre $(a, b)$

$\Rightarrow 2a = b$ and $b = 2 \Rightarrow a = 1$ and $b = 2$

$\therefore$ equation $x^2 + y^2 – 2x – 4y + c = 0$

Standard 11

Mathematics

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