The tangent to the curve $y=x^2+1$ at $(2,5)$ meets the normal to the curve at $(1, 2)$ 
Find the coordinates of the point where the tangent to the curve $y=x^2+1$ at the point $(2,5)$ meets the normal to the same curve at the point $(1,2).$

I tried to form 2 equations for each set of coordinates given, then solve then simultaneously to get the 'y and x' coordinates needed, however I didn't seem to get the answer right.
 A: *

*First, you need to find $f'(x)= 2x$. 

*Then to evaluate the slope of the tangent line at the point $(2, 5)$,
$m_1 = f'(2) = 4$. With slope $m_1=4$, and the point $(x_0, y_0) =
   (2, 5)$, use the slope-point form of an equation to obtain the
equation of that tangent line.


$(y - y_0) = m_1(x - x_0)\tag{Point-Slope Equation Of Line}$


*

*To find the slope of the normal line at the point $(1, 2)$, first
evaluate $f'(1) = m_2= 2$, the slope of the tangent line at that
point. Then the slope of the normal line to the curve at that same
point is $m_\perp = -\dfrac 1{m_2} = -\dfrac 12$. Using the slope of
the normal line $m_\perp = -\dfrac 12$ and the point $(x_0, y_0) =
   (1, 2)$, use the slope point form of an equation to obtain the
equation of the line normal to the curve at $(1, 2)$:
$$(y - y_0) = m_\perp(x - x_0)$$

*Finally, find where the tangent line and the normal line intersect by
setting the equations equal to one another. Solve for $x$, then find
the corresponding $y$ value using the equation of either line.
A: $y'(x) = 2x$
$y'(2) = 4$, so $4$ is the slope of the tangent at $(2,5)$.
You have a point and a slope, so you should be able to find the equation of a line.
Then $y'(1) =2$, so we invert and negate this to get the normal slope ($=-1/2$).
Point + slope --> equation.
Etc.
