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maths question help please

Just stuck on iv
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Reply 1
Avatar for oakjpw
oakjpw
1
The portion of the line between A and B will be a diameter. The center is therefore the midpoint. Hope that starts you off!
Reply 2
Avatar for Kalabamboo
Kalabamboo
OP
14
Original post by oakjpw
The portion of the line between A and B will be a diameter. The center is therefore the midpoint. Hope that starts you off!


But the portion of the line between A and B could possibly be a chord.
Reply 3
Avatar for mqb2766
mqb2766
21
The perpendicular bisector is the set of points equidistant from A & B. Find where this intersects with the x=3 line and that does the job?
(edited 5 years ago)
Reply 4
Avatar for oakjpw
oakjpw
1
Original post by Kalabamboo
But the portion of the line between A and B could possibly be a chord.


It can't be a chord because you are told the centre lies on that portion of the line. Any line joining two points of the circumference through the centre is a diameter.
Reply 5
Avatar for Kalabamboo
Kalabamboo
OP
14
Original post by mqb2766
The perpendicular bisector is the set of points equidistant from A & B. Find where this intersects with the x=3 line and that does the job?


Why specifically the intersection of the perpendicular bisector line with the line x=3, why not the intersection of the line AB with the line x=3?
The mark scheme has the same method as you I think -just dont understand this part
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(edited 5 years ago)
Original post by Kalabamboo
Just stuck on iv
e.PNG


Your equation is (x-3)^2 + (y-b)^2 = r^2
The two points a and b lie on the circle, so they satisfy the equation.
Create 2 simultaneous equations by plugging in the x and y coordinates for A and B and solve simultaneously.
Reply 7
Avatar for oakjpw
oakjpw
1
That is overcomplicating a lot for just 4 marks I think.
Reply 8
Avatar for Kalabamboo
Kalabamboo
OP
14
Original post by oakjpw
It can't be a chord because you are told the centre lies on that portion of the line. Any line joining two points of the circumference through the centre is a diameter.

"A circle passing through A and B has its centre on the line x = 3. Find the centre of the circle
and hence find the radius and equation of the circle"
The question doesn't say that the centre lies on the AB line. It just says that the circle passes through the points A and B and that the circle has its centre on the line x=3
Here is the mark scheme
f.PNG
Reply 9
Avatar for Kalabamboo
Kalabamboo
OP
14
Original post by mqb2766
The perpendicular bisector is the set of points equidistant from A & B. Find where this intersects with the x=3 line and that does the job?

Why specifically the intersection of the perpendicular bisector line with the line x=3, why not the intersection of the line AB with the line x=3?
The mark scheme has the same method as you I think -just dont understand this part
Original post by Kalabamboo
"A circle passing through A and B has its centre on the line x = 3. Find the centre of the circle
and hence find the radius and equation of the circle"
The question doesn't say that the centre lies on the AB line. It just says that the circle passes through the points A and B and that the circle has its centre on the line x=3
Here is the mark scheme
f.PNG


This is just a bit of loci. The perpendicular bisector have you a line of points which are the same distance from a and b. So the point where x = 3 has the same distance from a and b, much like a radius. So using the perpendicular bisector works here since it's not necessarily a diameter.
Original post by Kalabamboo
Why specifically the intersection of the perpendicular bisector line with the line x=3, why not the intersection of the line AB with the line x=3?
The mark scheme has the same method as you I think -just dont understand this part
f.PNG


My first reaction was to use quadratics and find r^2 for the center lying at (3.y) and then solve for y, but this is too much for 4 marks.

Just draw the picture of the line AB and the perpendicular bisector. All the points on the perpendicular bisectior must be the same distance from A & B, so the circle's center must lie on this line. If you're told the center also lies on x=3, then you simply substitute x=3 into the equation for the perpendicular bisector (intersection of the two lines).

Just draw the lines and remember you can construct the perpendicular bisector using a pair of compasses and drawing circles centered at A & B. Once you understand it graphically, it's fairly obvious. AB is a chord, the center must lie on the perpendicular bisector, something like https://www.bbc.com/education/guides/zsv8xfr/revision/7
(edited 5 years ago)
Original post by mqb2766
My first reaction was to use quadratics and find r^2 for the center lying at (3.y) and then solve for y, but this is too much for 4 marks.

Just draw the picture of the line AB and the perpendicular bisector. All the points on the perpendicular bisectior must be the same distance from A & B, so the circle's center must lie on this line. If you're told the center also lies on x=3, then you simply substitute x=3 into the equation for the perpendicular bisector (intersection of the two lines).

Just draw the lines and remember you can construct the perpendicular bisector using a pair of compasses and drawing circles centered at A & B. Once you understand it graphically, it's fairly obvious. AB is a chord, the center must lie on the perpendicular bisector, something like https://www.bbc.com/education/guides/zsv8xfr/revision/7

Thanks a lot:smile:

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