Question 1093746
<pre><font size = 4><b>
{{{drawing(2800/13,400,-2,5,-2,11,
locate(3,6,"P(3,6)"),locate(1,7.73,"O(1,h)"),locate(-1.8,1.3,"I(0,1)"),
graph(2800/13,400,-2,5,-2,11),  line(-9,-14,9,16),
circle(1,36/5,2sqrt(34)/5), line(3,6,1,36/5)  )}}}  

Comparing with the slope intercept form y=mx+b, the tangent line IP
whose equation is y=kx+1 has y-intercept (0,b) = I(0,1), and 
slope m=k.

Since the line goes through I(0,1) and P(3,6), we use the
slope formula to determine m = k, its slope:

{{{m=(y[2]-y[1])/(x[2]-x[1])}}}

{{{k=(6-1)/(3-0)=5/3}}}  <-- that's the answer for k

The radius OP is perpendicular to line IP, so its slope is
the reciprocal of {{{5/3}}} with the opposite sign, so the
slope of the radius OP is {{{-3/5}}}

To find the equation of OP, we use the point-slope form of 
a line with {{{m=-3/5}}} and the point of tangency, P(3,6).

{{{y-y[1]=m(x-x[1])}}}
{{{y-6=expr(-3/5)(x-3)}}}  <--equation of radius OP

Since center O(1,h) is on that radius, we substitute the
point (1,h), to find h

{{{h-6=expr(-3/5)(1-3)}}}
{{{h-6=expr(-3/5)(-2)}}}
{{{h-6=6/5}}}
{{{h=6+6/5}}}
{{{h=30/5+6/5}}}
{{{h=36/5}}}     <---that's the answer for h

Edwin</pre></font></b>