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Help with Circle question, A level C2.

Hi all,

I have been re-visiting old topics recently in prep. for the c2 exam. Today I am doing circles, but I can't do this question. I know I have done it before, but I have forgotten which makes it even more frustrating :s-smilie:

Show that the circle with equation + - 2ax - 2bx + = 0 touches the Y-axis.

Thanks a lot for any help :smile:
Reply 1
At the y-axis you have x=0

Set x=0 and see what happens
Reply 2
Original post by TenOfThem
At the y-axis you have x=0

Set x=0 and see what happens


Hi, I did that and got + = 0

But how does this show that it touches the y axis?

Thanks
Reply 3
I think that you have copied the question incorrectly and you should have -2by not -2bx
Reply 4
I don't think you can answer this question unless you have the values for a and b as they're the numbers that give you the translation from your original circle..
Reply 5
Original post by TenOfThem
I think that you have copied the question incorrectly and you should have -2by not -2bx


I double checked and the way I copied it is the way its in the book, but maybe this is a typo in the book? Out of interest, what would you do if it was -2by?
Reply 6
If it is -2by

x2+y22ax2by+b2=0x^2 + y^2 - 2ax - 2by +b^2 = 0


(xa)2a2+(yb)2=0(x-a)^2 - a^2 + (y-b)^2 = 0

(xa)2+(yb)2=a2(x-a)^2 + (y-b)^2 = a^2


When x=0, y=b ... a single solution so the circle touches the y-axis at (0,b)



So, yes, I think it is a typo in the book :smile:
Reply 7
Original post by TenOfThem
If it is -2by

x2+y22ax2by+b2=0x^2 + y^2 - 2ax - 2by +b^2 = 0


(xa)2a2+(yb)2=0(x-a)^2 - a^2 + (y-b)^2 = 0

(xa)2+(yb)2=a2(x-a)^2 + (y-b)^2 = a^2


When x=0, y=b ... a single solution so the circle touches the y-axis at (0,b)



So, yes, I think it is a typo in the book :smile:


Thats makes sense, thanks a lot! :smile:

Stupid textbooks =_=

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