Question

4.

Answer 1

Answer:

a) $(x-2)(x-6) + (y-5)(y-11) =0\Rightarrow (x-4)^{2}+(y+3)^{2}=68\\C(4,-3) \:r=\sqrt{68}$ 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:

$(x-a)^{2}}+(y-b)^{2}=r^{2}$

2) Let's expand the factored one into one closer to the pattern above:

$x^{2}-8x+12+y^{2}+6y-55=0\Rightarrow x^{2}-8x+y^{2}+6y=43$

3) Completing the square for both trinomials:

$(x-4)^{2}+(y+3)^{2}=43+16+9\Rightarrow (x-4)^{2}+(y+3)^{2}=68$

4) In the Reduced Formula, $(x-a)^{2}}+(y-b)^{2}=r^{2}$,

$C(a,b) \Rightarrow C(4,-3) \:the\:radius\:is\:r=\sqrt{68} \Rightarrow r=2\sqrt{17}$

b) Using the previous example to show this:

When we  factor this way

$(x-a)(x -c) + (y -b)(y-d) = 0$

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 $\sqrt{68} \cong 8.25$

So yes, the equation of the circle can be written as

$(x-a)(x -c) + (y -b)(y-d) = 0$

As long as (a,b) and (c,d) are the endpoints of of the diameter.

$d_{AC}=d_{BC}=R$