A circle passes through the point 3,$\sqrt{\frac{7}{2}}$ and touches the line pair $x^2-y^2-2x+1=0$. The co-ordinates of the centre of the circle are:-
My attempt:-
Using quadratic formula to separate the line pair into two lines.
$$x^2-y^2-2x+1=0$$
$$x=\frac{-(-2) \pm \sqrt{2^2-4(1-y^2)(1)}}{(2)(1)}$$
$$x=1\pm y$$
So we have two lines x-y-1=0 and x+y-1=0.I don't know that to do next?
Best Answer
So your two lines are $y=x-1$ and $y=-x+1$.
By symmetry we know the center of the circle must be on $x$-axis. Let the center be $(a,0)$ then the radius is $\large{a-1\over\sqrt{2}}$ because the angle between the two lines is 90 degrees and hence the two radius along with the two lines form a square where the $x$-axis is the diagonal.
Hence the circle is $(x-a)^2+y^2=\large{(a-1)^2\over2}$. Sub in your point, $(3-a)^2+{7\over2}=\large{(a-1)^2\over2}$.
$a^2-10a+24=0$ and you can find $a=4,6$.