#centre of circle with an unknown

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hardy dewBOT
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tepid basin
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Try completing the square.

burnt valley
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I tried

tepid basin
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What did you get?

burnt valley
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(x-1/2k)^2 - 0.25 + (y+2)^2-4 = sqrt(5)

zenith hemlock
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How’s k is in the equation

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Is it asking for k?

tepid basin
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You can only complete the square with y.

burnt valley
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Ohhh

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Ive got the y coordinate of the centre

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just not the x

tepid basin
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Well, as we just have x^2, the x-coordinate of the center is 0.

burnt valley
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tysm

tepid basin
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Did you get x^2 + (y + 2)^2 = k + 4?

burnt valley
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not even close

tepid basin
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Hm. Try again, then.

burnt valley
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yea i will thank you

burnt valley
tepid basin
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So, if we got x^2 + (y + 2)^2 = k + 4 and R = 5, what's the center and k?

burnt valley
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would the centre be (0,-2)?

tepid basin
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Yes.

burnt valley
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thanks for the help i rlly appreciate it

tepid basin
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Well, what about k?

burnt valley
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would k be -4

tepid basin
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No.

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Comparing the equations, we have:
k + 4 = R^2

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So, you can find k. Remember that we are given R = 5.

burnt valley
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OHH OK

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21 ty

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i acc suck at maths rn

tepid basin
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No worries! You just need some practice.

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Now, what about (b)?

burnt valley
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do i just sub coordinates into sqrt ((a2-a1)^2 + (b2-b1)^2)

tepid basin
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No, not yet.

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First, we need to find p.

burnt valley
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oh wow im blind

tepid basin
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To do that, substitute its coordinates into the equation of our circle.

burnt valley
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i get p^2 = -9 but that seems very wrong

tepid basin
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Yeah, that's not right.

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Can you show what you did?

burnt valley
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I did p^2 + 2^2 + 4(2) -21 =0 then tried to solve for p

tepid basin
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Well, that's correct. Not sure how you got p^2 = -9, though.

burnt valley
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Oh wait p=3 i think

tepid basin
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Yeah, you should get p^2 = 9. The positive root is p = 3.

burnt valley
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now do i just sub the coords into the formula

tepid basin
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Yes.

burnt valley
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alr ty

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I got 5 root 2

tepid basin
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Yeah, that's correct.