The centre of circle whose normals are x2 - 2xy - 3x + 6y = 0 i
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Actually, normal of any circle passing through it centre .
∴ first of all solve the given equation of normal , and get point , this is not other than centre of circle .
e.g., x² - 2xy - 3x + 6y = 0
⇒x(x - 2y) - 3(x - 2y) = 0
⇒(x - 3)(x - 2y) = 0
⇒ x = 3 and x = 2y
so, y = x/2 = 3/2
Hence, centre of circle is (3,3/2)
∴ first of all solve the given equation of normal , and get point , this is not other than centre of circle .
e.g., x² - 2xy - 3x + 6y = 0
⇒x(x - 2y) - 3(x - 2y) = 0
⇒(x - 3)(x - 2y) = 0
⇒ x = 3 and x = 2y
so, y = x/2 = 3/2
Hence, centre of circle is (3,3/2)
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