Answer:
a) [tex](x-2)(x-6) + (y-5)(y-11) =0\Rightarrow (x-4)^{2}+(y+3)^{2}=68\\C(4,-3) \:r=\sqrt{68}[/tex] b) (a,b) (c, d) As long as (a,b) and (c,d) are the endpoints of of the diameter.
Step-by-step explanation:
a)
1) The reduced formula of the Circumference is given by:
[tex](x-a)^{2}}+(y-b)^{2}=r^{2}[/tex]
2) Let's expand the factored one into one closer to the pattern above:
[tex]x^{2}-8x+12+y^{2}+6y-55=0\Rightarrow x^{2}-8x+y^{2}+6y=43[/tex]
3) Completing the square for both trinomials:
[tex](x-4)^{2}+(y+3)^{2}=43+16+9\Rightarrow (x-4)^{2}+(y+3)^{2}=68[/tex]
4) In the Reduced Formula, [tex](x-a)^{2}}+(y-b)^{2}=r^{2}[/tex],
[tex]C(a,b) \Rightarrow C(4,-3) \:the\:radius\:is\:r=\sqrt{68} \Rightarrow r=2\sqrt{17}[/tex]
b) Using the previous example to show this:
When we factor this way
[tex](x-a)(x -c) + (y -b)(y-d) = 0[/tex]
We are indeed, naming "a" and "b", the coordinates of (a, b) of the first endpoint and "b" and "d" the second endpoint as well.Id est, D (2, 5) and B (6,-11).
The radius, is [tex]\sqrt{68} \cong 8.25[/tex]
So yes, the equation of the circle can be written as
[tex](x-a)(x -c) + (y -b)(y-d) = 0[/tex]
As long as (a,b) and (c,d) are the endpoints of of the diameter.
[tex]d_{AC}=d_{BC}=R[/tex]
