No clue I’m not great at math but you will figure it ou
It is A because you are adding 3 to in if you were to subtract then it will go down not up.
So ASA is angle side angle, and that means that if you prove that the side, and the side adjacent to that side and the angle between those two sides are all congruent to another triangle's sides and angle, the triangles are both congruent.
The AAS is angle angle side, or something, so say you have a triangle and you prove that two of its angles are congruent along with a side to another triangle's, then it's AAS. I understand where the confusion might be. I guess it's just a matter of what you state first in your proof?
The answer is B! Good luck!
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)
