Question

a circle having its centre in the first quadrant touches the y axis at the point 0, 2 and passes through the point 1, 0 find the equation of the circle​

Answer 1

Equation of a circle is given by , when h and k are the centre of the circle and radius is r is :

(x-h)2 + (y-k)2 = r2

As it is given that the circle touches the y-axis, so h = r

Hence (x-h)2 + (y-k)2 =h2 (1)

And the circle is passing through (1,0 ) and (0,2) , hence both point will satisfy the equation

So (1-h)2 + (0-k)2 =h2

1 + h2 -2h + k2 = h2

Or 1 -2h + k2 = 0 (2)

And (0-h)2 + (2-k)2 = h2

Or 2+k2 -4k = 0

(k-2)2 = 0

Or k = 2 (3)

Putting (3) in (2), we get

h = 5/2

Hence putting the value in the standard equation, we have

(x -5/2) 2+(y-2)2 = (5/2)2

Or x2 + y2 -5x -4y + 4 = 0 is the required equation.

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Answer 2

hope this solution stands clear.