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Find the length of intercept made by the straight line $x+y=3$ with circle $x^2+y^2-2x-1=0$.

My Attempt:

Equation of circle.. $$x^2+y^2-2x-1=0$$ Comparing above equation with $$x^2+y^2+2gx+2fy+c=0$$

centre$=(-g,-f)$, $=(1,0)$.

Now, what should I.do next?

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    That's essentially the [same answer](http://math.stackexchange.com/questions/2129871/the-extremities-of-a-diameter-of-a-circle): determine the intercepts, then calculate the distance between them. For example, take $y=3-x$ from the first equation and substitute into the second one to get a quadratic in $x$. – 2017-02-05
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    @@dxiv, what do I do if I get the values of $x$? – 2017-02-05
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    Remember what you are trying to do: find the two intersections between the line and the circle. The quadratic in $x$ will give you the corresponding $x$ values. For each of them, you then know that $y=3-x$, so you have determined the two points of intersection. Now calculate the distance between them. – 2017-02-05
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    @ dxiv, solving this way i got; $$(x-2)^2=0$$ $$x=2$$. Only one point... – 2017-02-05
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    That's correct. The given line is tangent to the circle, so the length of the intercept is $0$. – 2017-02-05
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    @dxiv, The answer in my book is $2\sqrt {2}$ units. – 2017-02-05
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    If that's the right answer, then it's to a different question than what you asked here. Doublecheck that you didn't mistype the equations above. – 2017-02-05
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    @dxiv, I am sorry for the typo. its $x^2+y^2-2x-3=0$? – 2017-02-05
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    That's OK, now you know the procedure to solve it. – 2017-02-05

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Calculating the distance from centre $(1,0)$ of circle to the line $x+y=3$,

$d=\frac{|1+0-3|}{\sqrt{1^2+1^2}}=\sqrt 2=$radius of circle.

The answer must be zero as the line $x+y=3$ touches your circle.