Example Solution for Linear Equation:

Set BA = CA

BA = (x+1)^2 + (y-2)^2 = 5^2 (expand)

= (x + 1) (x + 1) + (y - 2) (y - 2) = 25

= x^2 + 2x + 1 + y^2 - 4y + 4 - 25 = 0

 

CA = (x + 6)^2 + (y - 6)^2 = 4^2 (expand)

= (x + 6) (x + 6) + (y - 6) (y - 6) = 16

= x^2 +12x + 36 + y^2 - 12y + 36 - 16 = 0

 

Now that both equations equal 0, we can set them equal to each other:

x^2 + 2x + 1 + y^2 - 4y + 4 - 25 = x^2 +12x + 36 + y^2 - 12y + 36 - 16

x^2 + 2x + y^2 - 4y - 20 = x^2 +12x + y^2 - 12y + 56 (subtract over to left side)

x^2 + 2x + y^2 - 4y - 20 - x^2 -12x - y^2 + 12y - 56 = 0 (combine like terms)

-10x +8y - 76 = 0 (add 10x + 76 to both sides)

8y = 10x + 76 (divide both sides by 8)

y = 1.25x + 9.5

 

On the GSP presentation, this is the equation for line AE (Show Line AE). This line intersects circle BA and circle CA at the points A and E.

 

Setting equation CA = DA produces the equation for line FA, y = - .75x - 2.5.

This line intersects circle CA and DA at points A and F.

 

Setting equation BA = DA produces the equation for line AC, x = -6.

This line intersects circle BA and DA only at one point, point A. This should be obvious since this is the only point of intersection for the two circles. Thus, the vertical line x = -6 is tangent to both circle BA and DA at point A.