A level Maths Circle Question

13

The question is...
Screenshot 2025-01-12 174638.png

A circle passes through the points A, B and C. Find the equation of the circle.

For context, l1 and l2 are perpendicular bisectors of their respective lines.
Previously, they got you to prove that D is (8,3). I didn't want too assume this was the centre of the circle. So I figured out that AC is a diameter, midpoint must be centre, and half the diameter is the radius, I got the equation (x-7)^2 + (y-5)^2 = 45.

In the answers, it said any perpendicular bisector of a chord passes through the centre of the circle. Is what I did correct? This is a question from MadasMaths not a past paper question. Also, is this circle theorem (in bold) one we need to know for the exam? I can't remember learning it but I might have forgot.

Reply 1

mqb2766

21

Original post

by Liqht

The question is... Screenshot 2025-01-12 174638.png

A circle passes through the points A, B and C. Find the equation of the circle.
For context, l1 and l2 are perpendicular bisectors of their respective lines.
Previously, they got you to prove that D is (8,3). I didn't want too assume this was the centre of the circle. So I figured out that AC is a diameter, midpoint must be centre, and half the diameter is the radius, I got the equation (x-7)^2 + (y-5)^2 = 45.
In the answers, it said any perpendicular bisector of a chord passes through the centre of the circle, so D is the centre. This is from MadasMaths, not a past paper, so I'[the question is... Screenshot 2025-01-12 174638.png

a circle passes through the points a, b and c. find the equation of the circle.
previously, they got you to prove that d is (8,3). i didn't want too assume this was the centre of the circle. so i figured out that ac is a diameter, midpoint must be centre, and half the diameter is the radius, i got the equation (x-7)^2 + (y-5)^2 = 45.
in the answers, it said any perpendicular bisector of a chord passes through the centre of the circle, so d is the centre. this is from madasmaths, not a past paper, so i'[]

AC would only be a diagonal if <ABC was a right angle (thales - gcse circle theorem). Madas is right, the perpendicular bisector of a chord passes through the center (simple isosceles triangle argument) so the intersection of two perpendicular bisectors gives the center, then sub a point (pythagoras) to get the radius (r^2)

Reply 2

the bear

20

the perpendicular bisector of any chord is a diameter

Reply 3

Liqht

OP

13

Original post

by mqb2766

AC would only be a diagonal if <ABC was a right angle (thales - gcse circle theorem). Madas is right, the perpendicular bisector of a chord passes through the center (simple isosceles triangle argument) so the intersection of two perpendicular bisectors gives the center, then sub a point (pythagoras) to get the radius (r^2)

I see, but is what I did wrong? Because the way I see it there's 2 different methods, his and mine, that give you two different centres?

Reply 4

Liqht

OP

13

Original post

by mqb2766

AC would only be a diagonal if <ABC was a right angle (thales - gcse circle theorem). Madas is right, the perpendicular bisector of a chord passes through the center (simple isosceles triangle argument) so the intersection of two perpendicular bisectors gives the center, then sub a point (pythagoras) to get the radius (r^2)

wait scratch my reply, I just understood thank you! :smile:

Reply 5

mqb2766

21

Original post

by Liqht

I see, but is what I did wrong? Because the way I see it there's 2 different methods, his and mine, that give you two different centres?

Yours is not correct as <ABC is not a right angle. Youve not put the values for A,B,C but it looks easy to verify that. If <ABC was right then the two methods would give the same result which is easy to verify (sketch, similar, a bit of trig)

Reply 6

Liqht

OP

13

Original post

by mqb2766

Yours is not correct as <ABC is not a right angle. Youve not put the values for A,B,C but it looks easy to verify that. If <ABC was right then the two methods would give the same result which is easy to verify (sketch, similar, a bit of trig)

Yeah I realised I'd wrongly assumed AC to be the diameter when it could just be another chord. Thanks for the help tho :smile:

Reply 7

mqb2766

21

Original post

by Liqht

Yeah I realised I'd wrongly assumed AC to be the diameter when it could just be another chord. Thanks for the help tho :smile:

Exactly, it is another chord and you could used it with one of the other ones to get the center.

Reply 8

mqb2766

21

Original post

by Liqht

Yeah I realised I'd wrongly assumed AC to be the diameter when it could just be another chord. Thanks for the help tho :smile:

Coming back to your original post, then knowing that a perpendicular bisector passes through the center is pretty much gcse. If you think about their basic definitions. so

•

A circle is the locus of points which are equidistant from a point (the center) pretty much defines the pythagoras circle equation x^2 + y^2 = r^2

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The locus of points which are equidistant from any two points is the perpendicular bisector of the line joining the points. This can be justified using greek construction so using a compass and draw arcs from each point and join the intersection points. Or note that each point on the locus is equidistant from the pair of points by drawing an isosceles triangle or ....

•

So if you have 3 points lying on the circumference, picking two will give a chord and the perpendicular bisector must pass through the center as the centre is equidistant from every point on the circumference. The perpendicular bisector line is the locus of all possible circle centres for those two points.

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Drawing 2 chords will give two intersecting diameters which defines the center as all 3 points are equidistant from the intersection point/center. Obviously, by definition/construction/... the perpendicular bisector of the 3rd chord (pair of points) would pass through this point.

Its covered at gcse, but most of the stuff is pretty elementary knowledge when thinking about what the lines, circles represent. So the equidistant properties of a line being the locus of points equidistant from two points and a circle being the locus of points being equidistant form a single point.

(edited 1 year ago)

A level Maths Circle Question

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