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Hey there
If possible, please try and include the diagram - really helps people skimming the forum and especially people that want to answer your questions, not everyone has the textbook in question! Just something to note for next time.
The method you suggest should work - we can look at your arithmetic for you if you still can't spot an error. However, this method is often redundant when there's a quicker method like the solution has.
Given distinct points \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) all lie on the same circle, you can set up two equations
then solve for h and k, which basically gives you the position vector for the centre.
Hi again guys lol
I've spent an extremely long time on this question (19 b) and i still can't quite get it.
Would appreciate any help!
Thanks
Only going to give the full working if you really really can't do it. That being said, here are a few hints that you might want to consider (they do go progressively, so look at them in order).
What do you notice about \(\triangle EFX\) and \(\triangle CDX\)?
Note that you are given that BE = AF = BC - is there a convenient way to link this to the fact that \(|\vec{AB}| = k|\vec{BC}|\)?
Can you set up an equation that uses one of EC and EX or FX and FD in terms of k?