Check the picture below, so the circle looks more or less like that one.
well, the center of it is simply the Midpoint of those two points, and its radius is simply half-the-distance between them.
![~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 3 -5}{2}~~~ ,~~~ \cfrac{ 5 + 9}{2} \right)\implies \left( \cfrac{-2}{2}~~,~~\cfrac{14}{2} \right)\implies \stackrel{center}{(-1~~,~~7)} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bmiddle%20point%20of%202%20points%20%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-5%7D~%2C~%5Cstackrel%7By_1%7D%7B9%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B3%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%20%5Cqquad%20%5Cleft%28%5Ccfrac%7B%20x_2%20%2B%20x_1%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20y_2%20%2B%20y_1%7D%7B2%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28%5Ccfrac%7B%203%20-5%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%205%20%2B%209%7D%7B2%7D%20%5Cright%29%5Cimplies%20%5Cleft%28%20%5Ccfrac%7B-2%7D%7B2%7D~~%2C~~%5Ccfrac%7B14%7D%7B2%7D%20%5Cright%29%5Cimplies%20%5Cstackrel%7Bcenter%7D%7B%28-1~~%2C~~7%29%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[3 - (-5)]^2 + [5 - 9]^2}\implies d=\sqrt{(3+5)^2+(-4)^2} \\\\\\ d=\sqrt{8^2+16}\implies d=\sqrt{80}\implies d=4\sqrt{5}~\hfill \stackrel{\textit{half the diameter}}{\cfrac{4\sqrt{5}}{2}\implies \underset{radius}{2\sqrt{5}}}](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-5%7D~%2C~%5Cstackrel%7By_1%7D%7B9%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B3%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7Bdiameter%7D%7Bd%7D%3D%5Csqrt%7B%5B3%20-%20%28-5%29%5D%5E2%20%2B%20%5B5%20-%209%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%283%2B5%29%5E2%2B%28-4%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%3D%5Csqrt%7B8%5E2%2B16%7D%5Cimplies%20d%3D%5Csqrt%7B80%7D%5Cimplies%20d%3D4%5Csqrt%7B5%7D~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bhalf%20the%20diameter%7D%7D%7B%5Ccfrac%7B4%5Csqrt%7B5%7D%7D%7B2%7D%5Cimplies%20%5Cunderset%7Bradius%7D%7B2%5Csqrt%7B5%7D%7D%7D)
Answer:
Step-by-step explanation:
Surface area = 2πr(h+r)
= 2 * 3.14 * 8*(8+12)
=2*3.14*8*20
= 1004.8 square unit
Answer:
Its a inequality
Step-by-step explanation:
YOUR WELCOME LOVE YOU
Answer:
1/3, 3/9, 4/12
Step-by-step explanation:
Answer:
where is the square at I cannot see it