Carrotsticks
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Re: 2012 HSC MX2 Marathon
Second Question
Observe the diagram. The two loci share the same y value for their centres. The green circle and the blue circle are the largest possible values of k for it to intersect with the red circle whilst having two solutions (well, strictly 1 solution, but we will be using an inequality).
Looking at the radius of the red circle (which is obviously 2 as defined by the locus), the green and blue circle can only intersect on either side of the red circle, in order to have 1 solution.
This occurs when x=1 and x=5.
However since the centre of the circle lies on the line x=0, we can safely say that these x co-ordinates will also be the radius of the blue/green circle.
But the radius of the blue/green circle is the value of k.
Hence![](https://latex.codecogs.com/png.latex?\bg_white 1 < k < 5)
Second Question
![](/proxy.php?image=http%3A%2F%2Fi1142.photobucket.com%2Falbums%2Fn613%2Fcarrotjono%2Fimage.jpg&hash=e58d7aa98caffb81837b4f73654ab982)
Observe the diagram. The two loci share the same y value for their centres. The green circle and the blue circle are the largest possible values of k for it to intersect with the red circle whilst having two solutions (well, strictly 1 solution, but we will be using an inequality).
Looking at the radius of the red circle (which is obviously 2 as defined by the locus), the green and blue circle can only intersect on either side of the red circle, in order to have 1 solution.
This occurs when x=1 and x=5.
However since the centre of the circle lies on the line x=0, we can safely say that these x co-ordinates will also be the radius of the blue/green circle.
But the radius of the blue/green circle is the value of k.
Hence
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