Question 938868
First draw out the picture without the fancy visuals to distract you. So just draw out 3 circles along with tangents for the first two circles like so

<img src = "http://i150.photobucket.com/albums/s91/jim_thompson5910/k370ln99s2m83265af5r-1_zps686f0bbd.png">

Notice how I made the segment OO' a <font color="red">red</font> segment and I made O'P a <font color="green">green</font> segment.
I've also added in segments OD and O'C shown in <font color="blue">blue</font>
In addition, I marked a new point R which is the intersection between the circle for the earth and the segment OO''


"the distance from the center of the sun to the center of the earth is approximately 93,000,000 miles"
So, OO'' = 93,000,000


"the diameter of the sun is 870,000 miles", so the radius is half that: 870,000/2 = 435,000
So, OD = 435,000


"the diameter of the moon is 2160 miles", so the radius of the moon is 2160/2 = 1080
So, O'C = 1080


"the diameter of the earth is 7920", so the radius of the earth is 7920/2 = 3960
So, RO'' = 3960
Using the segment addition postulate, we can say


OO'' = OR + RO''
93,000,000 = OR + 3960
93,000,000 - 3960 = OR
92,996,040 = OR
OR = 92,996,040


During lunar apogee, the moon is 252,700 miles away from the earth. This means the length of segment O'O'' is 252,700
Using the segment addition postulate, we know
OO' + O'O'' = OO''
Let y be the length of OO' and solve for y


OO' + O'O'' = OO''
y + 252,700 = 93,000,000
y = 93,000,000 - 252,700
y = 92,747,300


Therefore, the distance from the center of the sun to the center of the moon during a lunar eclipse and during a lunar apogee is roughly 92,747,300 miles.


Summary so far
OD = 435,000
O'C = 1080
RO'' = 3960
OO' = 92,747,300
OR = 92,996,040


Let's go back to the drawing. Focus on just triangle ODP and the smaller triangle O'CP and add in the measurements you see in the summary above (well with the exception of RO'' and OR)

<img src = "http://i150.photobucket.com/albums/s91/jim_thompson5910/k370ln99s2m83265af5r-2_zps115a8004.png">

Now let x be the length of O'P shown in <font color="green">green</font>. We need to find the length of x to help us solve this problem.


From the drawing, we see that OP is the sum of OO' and O'P. 
OP = OO' + O'P
OP = 92,747,300 + x


We have similar triangles, so we can set up a proportion and solve for x


(OD)/(OP) = (O'C)/(O'P)
(435,000)/(92,747,300 + x) = (1080)/(x)
435,000x = 1080(92,747,300 + x)
435,000x = 1080(92,747,300) + 1080x
435,000x = 100,167,084,000 + 1080x
435,000x - 1080x = 100,167,084,000
433,920x = 100,167,084,000
x = (100,167,084,000)/(433,920)
x = 230,842.28429


The length of O'P is approximately 230,842.28429 miles.
Use this to find OP


OP = OO' + O'P
OP = 92,747,300 + 230,842.28429
OP = 92,978,142.28429


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That's a lot of work, but we know these two important pieces of info

OP = 92,978,142.28429
OR = 92,996,040


If OP > OR, then P will be to the right of R and P will be inside the circle of the earth. 
However, above we see that OP is actually less than OR. So P will actually be located to the left of point R.


Therefore, the eclipse is NOT possible. No shadow from the moon is cast at all. The moon is simply too far away from the earth.



"How far is P from the center of the earth during lunar apogee?" It's asking for the length of PO'', so let's find that


OO'' = OP + PO''
93,000,000 = 92,978,142.28429 + PO''
93,000,000 - 92,978,142.28429 = PO''
21,857.7157099992 = PO''
PO'' = 21,857.7157099992


Question: How far is P from the center of the earth during lunar apogee?
Answer: Roughly 21,857.7157099992 miles


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