points (1,2),(3,-4),(5,-6) laying on the circumference of a circle,find the cordinate of centre?
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
Hello, anjana!
EB has the best solution.
Here's another approach . . . requiring more Thinking.
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: .$\displaystyle (11,\,2)$