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# C1 circles help Watch

1. 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?

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2. (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
3. (Original post by TeeEm)
drop a perpendicular from the centre to the x axis

I tried doing that, but I got stuck.
Do I form a triangle?
4. (Original post by grahamcrackers)

I tried doing that, but I got stuck.
Do I form a triangle?
Yep. Think about the properties of triangle ABC
5. (Original post by grahamcrackers)

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?
6. (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?
7. (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)
8. (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!)
9. (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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