Parabola and Circle problem : The parabola $y =x^2-8x+15$ cuts the x axis at P and Q. A circle is drawn .......... Problem : 
The parabola $y=x^2-8x+15$ cuts the x axis at P and Q. A circle is drawn through P and Q so that the origin is outside it. Find the length at a tangent to the circle from O. 
My approach : 
Since the parabola $y=x^2-8x+15$  cuts the x axis therefore, its y coordinate is zero, 
Solving the equation: $x^2-8x+15=0$ we get two points $(3,0)$ and $(5,0)$.
Now how to proceed further with these two points, please suggest.  thanks..
 A: Where is the centre of the circle - at some point 
$C=(4,a)$
What is the square of the radius of the circle: 
$r^2=a^2+1$
What is the square of the distance from the origin to the centre of the circle:
$OC^2=4^2+a^2$
Let $S$ be a point on the circle where the tangent from the origin touches it. We have a right-angled triangle with $OS^2+CS^2=OC^2$ and we know that $CS^2=r^2$
Can you finish it from there?
A: Let $T$ be the point of tangency. By the Power of a Point "tangent-secant" theorem
$$\begin{align}
|\overline{OT}|^2 &= |\overline{OP}|\;|\overline{OQ}| \\
&= 3 \cdot 5
\end{align}$$
so that
$$|\overline{OT}| = \sqrt{15}$$

Others have observed that, although there are infinitely-many circles through $P$ and $Q$, the length of the tangent segment from $O$ is always the same. Consequently, the set of all those points of tangency forms a circle around $O$, and this circle is said to be "orthogonal" to each member of the infinite family: it crosses each member at right angles. Such configurations have importance in more-advanced geometry.  
A: 
In the brown right-angled triangle, we can see that:$$r^2=y_c^2+1$$
In the right-angled triangle formed on the x-axis with OR as its hypotenuse, we see that:$$m^2=y_c^2+4^2=y_c^2+16$$
Finally, in the right-angled triangle ORT, we see that:$$l^2=m^2-r^2=(y_c^2+16)-(y_c^2+1)$$
Hopefully you can finish off from here.
A: Note that we can use the general theorem that if $OT$ is a tangent to a circle and $OPQ$ is a straight line meeting the circle at points $P$ and $Q$ we have $OT^2=OP\cdot OQ$
[this can be proved using similar triangles given that the angle between $OT$ and $TP$ is equal to the angle between $OQ$ and $TQ$ (given $OQ\gt OP$)]
