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Thanks!!!

3x + 3y  =  10          Multiply through by  -3 .

-9x - 9y  =  -30        This is the first equation.

-9x - 9y  =  -30        This is the second equation.

Notice that the equations are identical, so really we know only one equation.

There are two variables in that equation, so there are infinitely many solutions.

The circle is tangent to the y-axis at  (0, 2) , so the y-coordinate of the center must be  2  .

We know..

(x - h)² + (y - 2)²  =  r²      , where  h  is the x-coordinate of the vertex and  r  is the radius.

(0 - h)² + (2 - 2)²  =  r²     →​     h²  =  r²     →     h  =  ± r

(8 - h)² + (0 - 2)²  =  r²     →​     (8 - h)² + 4  =  r²

In the last equation, substitute  ± r  in for  h .

(8 - ± r)² + 4  =  r²

Since  r  is a distance,  r  =  4.25

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OK. I see now.

(Anyone can answer)

Pick a 4 digit number.

We need to see a diagram or -  at least -  know another point

x^2 + y^2  =  2x + 2y

x^2  - 2x  + y^2 - 2y  = 0

Complete the square on x and y  and we have that

x^2 - 2x + 1   +  y^2 -2y + 1  =  2

( x - 1)^2  +  (y - 1)^2  =  2

x will be maximized when  y = 1...so....

(x - 1)^2  =  2       take the positive square root

x - 1  =  √2

x =  √2 + 1

I know all of us could use a word of thanks

y =  (x - 2) / 2       (1)

x^2 + y^2   =  8    (2)

Sub (1)  into (2)

x^2  +   [  (x - 2) / 2 ]^2 =  8

x^2  + (1/4) (x^2 - 4x + 4 )  =  8

4x^2  + x^2 - 4x + 4 =  32

5x^2  - 4x - 28  =  0

(5x - 14) (x + 2)  =  0 

Setting each factor to 0 and solving for x we have that

x = 14/5      and  x   =  -2

The corresponding y coordinates are

y = (1/2) (14/5 - 2)  =  (1/2)(4/5)  = 4/10  = 2/5

y =  (1/2) (-2 - 2)  =  -2

So....the intersection  points are  ( 14/5, 2/5)  and  ( -2, -2) 

And the midpoint of this is  (  [ 14/5 -2] / 2 , [2/5 - 2] / 2 )  =

( [4/5]/2 , [-8/5]/2)  =   ( 4/10, -8/10)  =  (2/5, -4/5)

Hello michaelcai

This should be the answer

Tell me if I am wrong..