co-ordinate geometry

**anjana** · November 23rd 2006, 12:47 AM

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

**earboth** · November 23rd 2006, 01:09 AM

Originally Posted by anjana

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 $\mathbb{R}^2$ with the centre $M(m_1, m_2)$ is:
$(x-m_1)^2+(y-m_2)^2=r^2$

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

$(1-m_1)^2+(2-m_2)^2=r^2$
$(3-m_1)^2+(-4-m_2)^2=r^2$
$(5-m_1)^2+(-6-m_2)^2=r^2$

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

**Soroban** · November 23rd 2006, 07:08 AM

Hello, anjana!

EB has the best solution.
Here's another approach . . . requiring more Thinking.

Points $A(1,2),\:B(3,-4),\:C(5,-6)$ 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 $AB$ is $(2,-1)$ . The slope of $AB$ is: $\frac{-4-2}{3-1} = -3$
The equation of the perpendicular bisector of $AB$ is:
. . $y - (-1) \:=\:\frac{1}{3}(x - 2)\quad\Rightarrow\quad y\:=\:\frac{1}{3}x - \frac{5}{3}$

The midpoint of $BC$ is $(4,-5).$ The slope of $BC$ is $\frac{-6+4}{5-3}\:=\:-1$
The equation of the perpendicular bisector of $BC$ is:
. . $y- (-5) \:=\:1(x-4)\quad\Rightarrow\quad y \:=\:x - 9$

The perpendicular bisectors intersect when:
. . $\frac{1}{3}x - \frac{5}{3}\:=\:x - 9\quad\Rightarrow\quad\boxed{x\,=\,11}$

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

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

Thread: co-ordinate geometry

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