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# circles watch

1. hi

1) AB is diameter of a circle
A = (1,0) B = (3,0)
Find equation of the circle

2) Find equation of the circle passing through the points:
(0,0) (0,4) (6,0)

Thanks x
2. (Original post by jazzyj)
hi

1) AB is diameter of a circle
A = (1,0) B = (3,0)
Find equation of the circle

2) Find equation of the circle passing through the points:
(0,0) (0,4) (6,0)

Thanks x
erm
you sure thats the ony question?
looks a bit too simple to be true
anyways
the radius for question 1 is obviously 1
and then try finding the center
its the centre of the 2 points
thus

its
(x-2)² + y² = 1
i think...

now for q2
take A(0,0) , B(0,4) , C(6,0)

use the equation
x² + y² + 2gx + 2fx + c = 0

the equations are
c = 0 ----(1)
16 + 8f + c = 0 ----(2)
36 + 12g + c = 0 -----(3)

solve them all

you`ll get f as -2
g as -3

thus...equate them al in the original equation
x² + y² - 6x - 4x = 0

sorry if im wrong
3. the answer to q1 is supposed to be (x-1)(x-3) + y^2 = 0

and for q2, (x-3)^2 + (y-2)^2 = 13
4. (Original post by jazzyj)
the answer to q1 is supposed to be (x-1)(x-3) + y^2 = 0
that is the same as (x-2)^2 + y^2 = 1 , try expanding and you'll see
qusetion 2 is the same, if you factorise the answers given by MalaysianDude, you'll see that they are in the same form as the answers you want
5. ah thank you!
6. Alternative solution
============

Write the points as,
O(0,0), A(0,4), B(6,0)

Now, if O, A and B all lie on the circle then OA and OB are chords of the circle. That means that the perpindiculars to the mid-points of the chords pass through the centre of the circle.
The perpindicular passing thro' the mid-point of OA is the line y=2.
The perpindicular passing thro' the mid-point of OB is the line x=3.
Thes two lines, or perpindiculars, intersect at the point (3,2) - the centre of the circle!
Since the centre of the circle is the point C(3,2), and the origin, O, is a point on the circle, then pythag gives the radius², OC², as 3² + 2² = 13.
So, eqn of circle is,

(x-3)² + (y-2)² = 13
=============

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