Question

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Answer 1

number 5

Answer 2

how do you do it

Answer 3

Since we know this is a circle, you have to use the circle formula
\[(x-a) ^{2} + (y-b)^{2} = r ^{2}\] And because we are given 3 points, we'll have 3 circle equations. Then we solve the system to find the radius.

Answer 4

B is plug and chug with the answer from A, and C is just proportions

Answer 5

oh, I didn't realize its a circle. Well now its easier

Answer 6

nope, still confused. Can you walk through part a please?

Answer 7

Ok, so this is a long shot, but bear with me for a second. We know that our 3 equations are \[x ^{2}+y ^{2} = r ^{2}\]\[(x+10)^{2}+(y-1)^{2}=r ^{2}\]\[(x-16)^{2}+(y-2)^{2}=r ^{2}\]
We'll label them 1, 2, and 3 respectively.
Since we have all of our equations equal to r squared, we can eliminate a variable by setting 2 of the equations equal to each other, thereby eliminating one part of the system.

Personally, I paired equations 1 and 2 together for this one. So I had \[x ^{2}+y ^{2} = (x ^{2}+20x+100)+(y ^{2}-2y+1)\]
Eliminate both x and y squared, and you're left with
0 = 20x+100-2y+1

I solved for y, which was y=10x+101/2

Now, I set equations 1 and 3 equal to each other, but this time, plugging in the y value.
\[(x ^{2})+(100x^{2}+1010x+\frac{ 10201 }{ 4 })=(x ^{2}-32x+256)+(100x ^{2}+970x+\frac{ 9409 }{ 4 })\]

Combined like terms and solving for x yields 72x = 58, x = 29/36

Now I can go back and solve for y this way by plugging x back in because it has an actual value. So I set 1 and 2 equal to each other. \[(29/36)^{2}+y ^{2}=(389/36)^{2}+(y ^{2}-2y+1)\]

Cleaning it up a bit
\[\frac{ 841 }{ 1296 }=\frac{ 152617 }{ 1296 }-2y\]

\[y=\frac{ 75888 }{ 1296 }\]

Finally, we can substitute in both values to find the radius
\[(\frac{ 29 }{ 36 })^{2}+(\frac{ 75888 }{ 1296 })^{2}=r ^{2}\]

\[3429.402006=r ^{2}\]
\[r = 58.561\]

Answer 8

I hope this helps

Answer 9

yes it did

Answer 10

@ixawesomeness this is your answer