points (1,2),(3,-4),(5,-6) laying on the circumference of a circle,find the cordinate of centre?:confused:

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- Nov 23rd 2006, 12:47 AManjanaco-ordinate geometry
points (1,2),(3,-4),(5,-6) laying on the circumference of a circle,find the cordinate of centre?:confused:

- Nov 23rd 2006, 01:09 AMearboth
Hello, Anjana,

I assume that the equation of a circle in $\displaystyle \mathbb{R}^2$ with the centre $\displaystyle M(m_1, m_2)$ is:

$\displaystyle (x-m_1)^2+(y-m_2)^2=r^2$

Now plug in the values yxou know. You'll get 3 equations:

$\displaystyle (1-m_1)^2+(2-m_2)^2=r^2$

$\displaystyle (3-m_1)^2+(-4-m_2)^2=r^2$

$\displaystyle (5-m_1)^2+(-6-m_2)^2=r^2$

Expand the LHS of these equations. Then substract equ2 from equ1 and equ3 from equ1. You'll get 2 linear equations without any squares which will give you the values of m1 and m2. Afterwards you can calculate the value of r.

For confirmation only M(11, 2) and r = 10

EB - Nov 23rd 2006, 07:08 AMSoroban
Hello, anjana!

EB has the best solution.

Here's another approach . . . requiring more Thinking.

Quote:

Points $\displaystyle A(1,2),\:B(3,-4),\:C(5,-6)$ are on the circumference of a circle.

Find the coordinates of the centre.

The circumcenter is the intersection of the perpendicular bisectors of the sides.

The midpoint of $\displaystyle AB$ is $\displaystyle (2,-1)$. The slope of $\displaystyle AB$ is: $\displaystyle \frac{-4-2}{3-1} = -3$

The equation of the perpendicular bisector of $\displaystyle AB$ is:

. . $\displaystyle y - (-1) \:=\:\frac{1}{3}(x - 2)\quad\Rightarrow\quad y\:=\:\frac{1}{3}x - \frac{5}{3}$

The midpoint of $\displaystyle BC$ is $\displaystyle (4,-5).$ The slope of $\displaystyle BC$ is $\displaystyle \frac{-6+4}{5-3}\:=\:-1$

The equation of the perpendicular bisector of $\displaystyle BC$ is:

. . $\displaystyle y- (-5) \:=\:1(x-4)\quad\Rightarrow\quad y \:=\:x - 9$

The perpendicular bisectors intersect when:

. . $\displaystyle \frac{1}{3}x - \frac{5}{3}\:=\:x - 9\quad\Rightarrow\quad\boxed{x\,=\,11}$

Then: .$\displaystyle y \:=\:11 - 9\quad\Rightarrow\quad\boxed{y \,=\,2}$

Therefore, the center of the circle is: .$\displaystyle (11,\,2)$