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C1 circles help

Hello everyone

I don't know if I'm being mindless as it is so late but for the life of me I do not know the method to find B in this question.

It does not give me an equation.

Can anyone help me?

Reply 1
Original post by grahamcrackers
Hello everyone

I don't know if I'm being mindless as it is so late but for the life of me I do not know the method to find B in this question.

It does not give me an equation.

Can anyone help me?



drop a perpendicular from the centre to the x axis
Original post by TeeEm
drop a perpendicular from the centre to the x axis


Thanks for the reply!

I tried doing that, but I got stuck.
Do I form a triangle?
Original post by grahamcrackers
Thanks for the reply!

I tried doing that, but I got stuck.
Do I form a triangle?


Yep. Think about the properties of triangle ABC
Reply 4
Original post by grahamcrackers
Thanks for the reply!

I tried doing that, but I got stuck.
Do I form a triangle?


what is the x coordinate of the point where your perpendicular meets the x axis?
Original post by TeeEm
what is the x coordinate of the point where your perpendicular meets the x axis?


I got B to be (7, 0)

I knew that A to C was moved by the vector across 3 and up to, so when I reached the centre I went down 2 and across 3.

Is this right? If so, I feel like this is a stupid way of doing it. Is there a nicer way?
Reply 6
Original post by grahamcrackers
I got B to be (7, 0)

I knew that A to C was moved by the vector across 3 and up to, so when I reached the centre I went down 2 and across 3.

Is this right? If so, I feel like this is a stupid way of doing it. Is there a nicer way?


you are correct

nicer I am not sure ...

there is a circle theorem which states that the perpendicular bisector of a chord always passes though the centre.

the chord here is the x axis
bisector cuts x axis at (4,0)
as this is the midpoint (perpendicular bisector) the other end must be at (7,0)
(edited 9 years ago)
Original post by TeeEm
you are correct

nicer I am not sure ...

there is a circle theorem which states that the perpendicular bisector of a chord always passes though the centre.

the chord here is the a axis
bisector cuts x axis at (4,0)
as this is the midpoint (perpendicular bisector) the other end must be at (7,0)


Ah!! I understand! This method is much easier.

Thank you for your time! I need to figure out how to give you rep! (i'm new here!)
Reply 8
Original post by grahamcrackers
Ah!! I understand! This method is much easier.

Thank you for your time! I need to figure out how to give you rep! (i'm new here!)


my pleasure

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