A lamp with height (h) and some other things, so find (h) "If a lamp with height ($h)$ was at  point $(-2,h)$, and the light ray that comes from it touches circle $(x^2 +y^2 = 1)$ at a point in the first quarter, and then stops when it hits the x axis at point $(1.25,0)$, so find the height ($h$)".
What I have understood is:
1.A triangle with sides ($h$, the x axis from -2 to 1.25 and the light ray) forms.
2.the light ray touches the circle in one point so i think we need the derivative somehow
3.if the light ray is going from
x = -2 to x = 1.25 and it is going from $y = h$ (which is obviously positive) to $ y = 0$, then it is going to go through the y axis at some point which i we dont know.
What do i want? just (h)
 A: Points are labelled as shown.


*

*Using the right-angled triangle SAD to find $\angle SAD$ which is the same as $\angle BCD$ (because ABCD is a cyclic quadrilateral).


*Find $\angle XSD$ which is $90^0 + \angle BCD$


*Use point-slope form to get the equation of SDC.


*Solve x = -2 and the equation in step 3 to get h.
Almost there.
A: The line segment $A(-2, h)$ to $(1.25, 0)$ is tangent to the circle $x^2 + y^2 = 1$.  Therefore the distance between the origin (which is the center of the circle) and the line segment is $1$.
Now, we just need to find the equation of the line segment, and this is given by
$ y = \dfrac{ h }{-2 - 1.25 } (x - 1.25) $
Cross multiplying yields
$ -3.5 y = h (x - 1.25) $
The distance from the origin $(0,0)$ to this line is given by
$ d = \dfrac{ | -3.5(0) - h( 0 - 1.25) | }{\sqrt{3.5^2 + h^2 } } = \dfrac{1.25 h}{\sqrt{3.5^2 + h^2 } } $
Since $d = 1$, then
$ 1.25^2 h^2 = 3.5^2 + h^2  $
From which
$ h^2 = \dfrac{3.5^2} { \bigg(\dfrac{9}{16}\bigg) }$
Taking the square root
$ h = \dfrac{ \bigg(\dfrac{7}{2} \bigg)}{ \bigg(\dfrac{3}{4}\bigg) } = \dfrac{14}{3} $
