+0  
 
0
102
2
avatar+777 

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

 Dec 9, 2018
 #1
avatar+98102 
+2

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)

 

 

cool cool cool

 Dec 9, 2018
edited by CPhill  Dec 9, 2018
 #2
avatar+777 
0

Ah ok i get it now, i wrote MR= 20 thanks!

YEEEEEET  Dec 9, 2018

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