Examination Question (1 Viewer)

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Can anyone do this question (it's a last question in the exam).
Question: Screen shot 2015-08-18 at 5.38.38 PM.png
Answer: Screen shot 2015-08-18 at 5.38.44 PM.png

I'd be grateful if anyone can help me, I think I can find what C1 is but having troubles finding C2
 

Kaido

be.
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2015
First let's re-write in general form:
(x-3)^2 + (y+k)^2=k^2-3k+9

For C1: Centre = (3,-k). Sub this into x-3y=0 -> yielding k=-1
For C2: Centre = (3,-k). Therefore we want to find y-coordinate. But y-coord is also = radius (as the circle touches the x-axis)
Therefore k=Sqrt (k^2-3k+9)
Square both sides and solve for k: k=3
 
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