More Maths Questions

[CoD]

Circle questions

a) Find the equation of the circle which passes though the points A (0,0) B(3,9) C(6,8)

The answer is x^2+y^2-6x-8y=0 but why? Can anyone even put me on the right track, of what to do?

b) Find the equation of the tangent to this circl at the piont C

Same again, I guess since its tangents you differentiate or integrate or something, but what, ne ideas?

Cheers

[Azof]

I'm on break, and it's way too early (CST) to be doing math so I'll just send a few pointers-

Complete teh square

Relationship of h [ (x-h)^2 ] and k [ (y-k)^2 ] to the vertex (h,k)

oh, and for the tangent thing, just use the radius as a distance from the line the circle is supposed to be tangent to, and construct the equation using your newly found vertex (st equation for a circle is (x-h)^2 + (y-k)^2 = r^2

any of this stuff is liable to be wrong as I was out all night

*edit*

distance formula is your friend... don't make it harder than it is

[midge5]

a) no idea

b) The centre of the circle it (3,4)
The gradient of this line to point C is 4/3

the gradient of the tangent x gradient at Centre = -1
gradient of tangent = -1/.(4/3) = -3/4

then put the info into the equation of a straight line

(Y-Y1) = M(x-x1)
(Y-6) = 3/4(x-8)
then just finish working it out for the equation of the tangent

[queball]

Originally posted by CoD
Circle questions

a) Find the equation of the circle which passes though the points A (0,0) B(3,9) C(6,8)

The answer is x^2+y^2-6x-8y=0 but why? Can anyone even put me on the right track, of what to do?

The gradient of AB is 3, the gradient of BC is -1/3. This should tell you that A(B)C is a right angle, so AC is a diameter. The rest is easy.

[queball]

Originally posted by midge5
a) no idea

b) The centre of the circle it (3,4)
The gradient of this line to point C is 0.75

4/3

[midge5]

Originally posted by queball
4/3

heh so it is :/ read the x coords as Y

['|'empest]

Use the three points to work out the equations of 2 lines passing through two points of the circle, The perpendicular bisectors of each of these lines will intersect at the centre of the circle, then you can use x^2 + y^2 = z^2 to work out the radius.

[W]

If you have three points [0,0] [a,b] and [c,d] then the center of the circle intersecting the three points lie at
[(b(c^2+d^2)-d(a^2+b^2))/(2(bc-ad)),(a(c^2+d^2)-c(a^2+b^2))/(2(ad-bc))]
Which can be calculated faster by precalculating:
o1=a^2+b^2
o2=c^2+d^2
o3=2(ad-bc)
it's then [(do1-bo2)/o3,(ao2-co1)/o3]
Since one of the points is origo, the radius calculation is trivial.