# Finding the center of circle touching a line pair and a point.

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?

• Hint: if a circle touches two lines, then its center lies on their angle bisector – shardulc says Reinstate Monica Nov 11 '15 at 3:41
• Also, if you want an equation for the circle (as opposed to a geometric construction), then once you have the center on the angle bisector, you can substitute the coordinates of the point into the equation and get a quadratic, which means there are two possible circles. – shardulc says Reinstate Monica Nov 11 '15 at 3:45

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 $a=4,6$.