Find the intersection of a vertical line segment in a circle. My brother needs help coming up with a formula for a problem that I already did but failed to write out the formula for. 
The problem is:
Consider a circle with the point (5,4) and a radius of 3.
Determine if a vertical line segment with the points (3,5) being one end of the segment and (3,1) being the other end of the segment, intersects this circle. List the intersecting points if so.
:
Again I need help coming up with a simple formula to determine where the line intersects the circle.
Thanks
 A: The points on the line segment each satisfy an equation and two inequalities, 
$$ x=3 \qquad y>1 \qquad y<5. $$
The points on the circle each satisfy the equation, 
$$ (x-5)^2+(y-4)^2=9. $$
An intersection of these curves will be an ordered pair, $(x,y)$, which solves both equations and satisfies the inequalities. 
Since every point on the vertical line segment has the x-coordinate $x=3$ we will start by substituting $x=3$ into the equation of the  circle.
$$ (3-5)^2+(y-4)^2=9 $$
$$ 4 + (y-4)^2 = 9 $$
$$ (y-4)^2=5 $$
$$ y-4=\pm\sqrt5 $$
$$ y=4\pm\sqrt5 $$
$$ y \approx 4 \pm 2.23 $$
$$ y\approx 1.77, 6.23$$
This gives us two ordered pairs $(3,1.77)$ and $(3,6.23)$. These both solve the equations for the circle and the line segment, but only one of them satisfies the inequalities. Since $1<y<5$, only the point $(3,1.77)$ is a valid point of intersection. 
A: The equation for the circle is:
$$
(x - 5)^2 + (y-4)^2 = 9
$$
The equation for the line is:
$$
(3, t) \ \ \ \ t\in[1,5].
$$
This gives:
$$
(3-5)^2 + (t-4)^2 = 4 + t^2 - 8t + 16 = 9 \\
\Rightarrow t^2 -8t + 11 = 0
$$
Using the quadratic formula to solve for $t$ gives:
$$
t = 4 -\pm  \sqrt{5}.
$$
Since $4 + \sqrt5 > 5$, the only valid value of $t$ is $4 - \sqrt5$, which, after plugging back into the equation for the line, gives the single intersection point:
$$
(3, 4 -\sqrt5)
$$
