# Find m and n if C(m,n) is the center of the circle circumscribed to the triangle AOB. A(4,-2), B(2,4)

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The formula for the circle is:

x^2 + y^2 + ax + by +c =0

We have the triangle AOB , then A ,O and B are on the circle, Then A(4,-2) , O(0,0) and B(2,4) should verify the equation:

First we will subtitute with O(0,0):

==> 0+0+0+0+c= 0

==> c=0

Now substitute with A(4,-2):

16 + 4 + 4a -2b = 0

20 + 4a -2b =0 ......(1)

Now substitute with B(2,4):

4+ 16 +2a +4b =0

20 + 2a +4b =0 .....(2)

Now let us multiply (1) with 2 and add to (2):

40 +8a -4b =0

20 + 2a +4b= 0

==> 60 + 10a =0

==> a= -60/10 = -6

==> 2b= 20 +4a= 20-24= -4

==> b= -2

Now substitute a and b in the equation:

x^2 + y^2 -6x -2y =0

x^2 -6x +y^2 -2y =0

Complete the squares:

(x-3)^2 + (y-1)^2 -9 -1 =0

(x-3)^2 + (y-1)^2 =10

Then the center C is (3,1).

If the circle is circumscribed to the triangle AOB, that means that the vertex A,B,O are on the circle and their coordinates verify the equation of the circle.

The general equation of the circle is:

x^2 + y^2 + ax + by + c = 0

Now, we'll substitute the coordinates of A,B and O, by their values, in the equation of the circle:

xA^2 + yA^2 + a*xA + b*yA + c = 0

16+4+4a-2b+c = 0 (1)

xB^2 + yB^2 + a*xB + b*yB + c = 0

4+16+2a+4b+c = 0 (2)

xO2 + yO^2 + a*xO + b*yO + c = 0

c=0

If c = 0, (1) and (2) will become:

20 + 4a -2b = 0 (3)

20 + 2a + 4b = 0 (4)

We'll divide by 2,(4).

4a -2b = -20 (5)

a+2b = -10 (6)

We'll add (5) and (6)

5a = -30

We'll divide by 5:

a = -6

We'll substitute a in the relation (6):

-6 + 2b = -10

We'll add 6 both sides:

2b = -4

We'll divide by 2:

b = -2

The equation of the circle is:

x^2 + y^2 - 6x - 2y = 0

We'll form the squares:

x^2 - 6x + 9 + y^2 - 2y + 1 - 9 - 1 = 0

(x-3)^2 + (y-1)^2 = 10

**The coordinates of the center of the circle are: C(3,1).**