Find the centre of the circle passes through (6,2), (0,4) & (4,6).

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Solution

Given that the circle passes through (6,2), (0,4) & (4,6)
Let the centre be O(x,y)

oa =ob =oc
$O A^{2} = (6 - x)^{2} + (2 - y)^{2} (i) O B^{2} = (0 - x)^{2} + (4 - y)^{2} (i i) O C^{2} = (4 - x)^{2} + (6 - y)^{2} (i i i)$
Equating (i) & (ii)
$(6 - x)^{2} + (2 - y)^{2} = (0 - x)^{2} + (4 - y)^{2} 36 + x^{2} - 12 x + 4 + y^{2} - 4 y = x^{2} + 16 + y^{2} - 8 y 4 y - 12 x + 24 = 0 y - 3 x + 6 = 0 - (i v) E q u a t i n g (i i) & (i i i) (0 - x)^{2} + (4 - y)^{2} = (4 - x)^{2} + (6 - y)^{2} \Rightarrow x^{2} + 16 + y^{2} - 8 t = 16 + x^{2} - 8 x + 36 + y^{2} - 12 y y + 2 x - 9 = 0 (v) y - 3 x + 6 = 0 y + 2 x - 9 = 0 - -------------- - 5 x = 15 x = 3 y = 3$
Hence, (3,3) is centre of circle

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