+845 
For a) i did -9+15 8+-10
which gave me 6,-2
divide that by 2 my answer is (3,-1)
B) from a this is what i have (x-3)^2 + (y+1)^2 = r^2
i did \(\sqrt{(3+9)^2 + (-1-8)^2}\)
which gave me \( \sqrt{202} \)
so my answer is (x-3)^2 + (y+1)^2 = 202
i just need help with c
thanks
+129852 OK, YEEEEEET....I think the radius should be sqrt 225 = 15
Remember that if we draw a perpendicular to the chord from A....this perpendicular will bisect the chord.......call the point where this perpendicular intersects the chord, M
And we will have a right triangle AMR
MR = 10
AR = 15 = the hypotenuse of AMR
The other leg, AM....is the distance we seek
So
AM = sqrt [ (15 )^2 - 10^2 ] = sqrt [ 225 - 100] = sqrt (125) = 5sqrt(5)
