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C1 Co ordinate geometry

username1533709

I'm stuck on part 2.The pictures are of my previous working.

Reply 1

smsgd

Original post by Kadak

I'm stuck on part 2.The pictures are of my previous working.

I can't see the pictures anywhere?

Reply 2

username1533709

Original post by smsgd

I can't see the pictures anywhere?

My app seems to be done.I'm not sure how to upload pics on the desk version but the question I'm stuck on is 9(II) on OCR paper 18 may 2011.

Original post by Kadak

Okay, so did you workout the right angle to be at A, B or C? (I cant quite see the top of the picture)

Original post by smsgd

Okay, so did you workout the right angle to be at A, B or C? (I cant quite see the top of the picture)

The right angle is at c I believe.

Reply 8

smsgd

Original post by Kadak

The right angle is at c I believe.

I got the right angle to be at A?

(edited 9 years ago)

Reply 9

username1533709

Original post by smsgd

I got the vertex to be at A?

Oh sorry,I thought It would be c cause it's opposite to it but yes the vertex is at A.

Reply 10

smsgd

Okay, so in that case, we know that BC is the diameter of the circle (due to the 'angle in a semicircle theorem' - if A is the right angle then the remaining to sides must join to give the diameter).

If BC is the diameter, the midpoint of the line is the centre of the circle. the coords are:
x: [7+ (-3)]/2 = 2
y: [1 + (-9)]/ 2 = -4

Lenght of BC is sqrt{[7-(-3)]²+ [1-(-9)]^2} = sqrt(200) = 10*sqrt(2), thereforce radius of the circle is 5*sqrt(2) (Half of BC which is the diameter)

The equation of the circle is therefore : (x-2)²(y+4) ²= [5* sqrt(2)]^2

Expand that and you should get the answer in the format the wuestion asks

(edited 9 years ago)

Reply 11

Arjun24

Not sure if this is clear enough, but this is what I just did. Hopefully its correct and it helps.

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1429397652037.jpg41.2KB

Reply 12

smsgd

Original post by Arjun24

Not sure if this is clear enough, but this is what I just did. Hopefully its correct and it helps.

Posted from TSR Mobile

If A is the right angle, wouldn't that make BC the diameter?

Reply 13

samb1234

Original post by smsgd

Okay, so in that case, we know that BC is the diameter of the circle (due to the 'angle in a semicircle theorem' - if A is the right angle then the remaining to sides must join to give the diameter).

If BC is the diameter, the midpoint of the line is the centre of the circle. the coords are:
x: [7+ (-3)]/2 = 2
y: [1 + (-9)]/ 2 = -4

Lenght of BC is sqrt{[7-(-3)]²+ [1-(-9)]^2} = sqrt(200) = 10*sqrt(2), thereforce radius of the circle is 5*sqrt(2) (Half of BC which is the diameter)

The equation of the circle is therefore : (x-2)²(y+4) ²= 5* sqrt(2)

Expand that and you should get the answer in the format the wuestion asks

Yeah this is the best way of doing this. You could setup a 3 way simultaneous equation but without a calculator that is very hard to do

Posted from TSR Mobile