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The equation of the circle, which touche...

The equation of the circle, which touches the parabola `y^2=4x` at (1,2) and passes through the origin is :

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The correct Answer is:
`x^(2)+y^(2)-7x+y=0`
The equation of the tangent at (1,2) to `y^(2)=4x` is x-y+1=0. let the equation of the required circle be `(x-1)^(2)+(y-2)^(2)+l(x-y+1)=0` it also passes through (0,0)
`implies l=-5`
Equation of required circle is `x^(2)+y^(2)-7x+y=0`.
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