Education What am I doing wrong? (Maths)

I need to find the equation of a circle with diameter AB where A=(-3,-6) and B=(9,2).
Please point the mistake(s) in my method/working.

So the first thing i did was find the midpoint of AB (which is also the centre of the circle) which came out to be (3,-2).
Then (this is where i think i went wrong as i got a surd which i don't think i should have gotten) I got the length of AB=sqrt{ (x2-x1)^2 + (y2-y1)^2 } = sqrt{ (9+3)^2 + (2+6)^2 } = 4 sqrt{13}
Then i halved this to get the radius 2 sqrt{13} and substituted the values into the general equation of a circle.

Pretty sure my mistake is somewhere in the above. Any ideas?

 

seems right to me :S

 

I think that's correct...

Gives you an end formula of (x-3)² + (y+2)² = 52, right?


(edit) you can use the [ sup ] tag for the exponents. (/edit)

 

Your numbers are definitely right.

This equation: (x - h)^2 + (y - k)^2 = r^2 ?
(x-3)^2 + (y+2)^2 = (2rt13)^2 = 52

I can't see anything wrong. I've ever got up this funky java circle plotter (http://www.analyzemath.com/CircleEq/CircleEq.html and click the button) and it's coming out fine.

 

It is right
Turns out the last part--simplifying the equation--which i was sure i had done correctly was the problem. I got some signs mixed up.
Thanks for all your help

 

Hehe, vector geometry is a pain in the ass - I'm so glad we don't do it anymore.

 

as someone who did advanced 3d graphics at university (and passed with flying colours, i might add! *trumpet blowage*)-

it's actually quite pointless.

 

Thanks for all your help on that but I have another question which I have no idea how to do.
There are 3 points: P(-2,7), Q(2,3), and R(4,5). The question first asks me to show that PQ is perpendicular to QR which I have done. It then asks for the equation of the circle which passes through the 3 points. I have no idea how to do the second part. Can anyone help me out? Just point me in the right direction if you can.
Thanks in advance

 

Try drawing them on a graph - you want to find a circle with the appropriate radius and centre that it fits all the points on the circumference. Diagrams always helped me.

 

I can't say a diagram's helping me in this case. It seems to me that neither PQ or QR has to be the radius of the circle so all i know is that they have to be on the circle but that's not enough to get the equation

 

Hmm... I see what you mean there - sadly my idea of having a diagram hasn't helped... If anything it just shows there's no conclusive point for the centre (that I can see off-hand), and no obvious radial direction.

 

Consider the general solution for a circle:
x^2+y^2-2ax-2by+c=0 where c=a^2+b^2-r^2

(x,y) is the cartesian co-ords of any point on the circle.
r is the radius.
a and b are constants you are trying to find.

Sub in x and y for your first point, and you'll get an linear equation in terms of a, b and c. Sub in x and y for your second point and you'll get another one. Do it again for your third one.

You'll have 3 linear equations in terms of a, b and c. Solve them simultaneously. Problem solved hope that helps.

 

I think it's easier to use (x-a)² + (y-b)² = r² than the multiplied-through version, but dragontail has the right approach.

 

or you can take two lines AB and BC, find their perpendicular bisectors, intersect those, use that as the center, and find the distance from the center to one of the points as the radius.

That is if you're more geometrically minded than algebraic.

 

dragontail what exactly do i do when i substitute the values into the equation. For example, for the point P(-2,7) i would substitute giving me:
x^2+y^2+4x-14y+c=0 correct?
Now what?

 

just looking at the question,

instinctively.. i think it's a square in a circle problem.

mid-points of 2 lines is the coordinate of the circle center?

 

Sub in x=-2 and y=7
You get:
4+49+4a-14b+c=0
which is rearranged to give
4a-14b+c=-53

Now do the same for the other two points and solve simultaneously. If you get no solution for the simultaneous equation, then the points are colinear.