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    Find in cartesian form, an equation of the circle, which touches the y-axis at the point (0,6) and also passes thru the point (1,3).

    Well I know the centre must be (something, 6), but hmm I'm stuck....I bet I'm missing something really simple! Hints would be appreciated
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    Let the centre by (x, 6). Then

    x^2 = (x - 1)^2 + (6 - 3)^2 = x^2 - 2x + 10
    => x = 5.
 
 
 

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