Question 1209063
<pre>

The green line is the set of all points having a y-coordinate of 6.
The green circle, with center (1,2) and radius 17, is the set of all points
whose distance from point (1,2) is 17.  So the 2 points they have in common
are the only two points having both properties.  So the answer will be those
two points.

B is 6 units above the x-axis and A is 2 units above the x-axis, so the
distance AB = 6-2 = 4.

We have AB=4 and AC=17, {{{BC=sqrt(AC^2-AB^2)=sqrt(17^2-4^2)=sqrt(289-16)=sqrt(273)}}}.

B is 1 unit farther to the right of the y-axis, so is C, and therefore

{{{C=(matrix(1,3,1+sqrt(273),",",6))}}}

That makes C' one unit less to the left of the y-axis, so

{{{"C'"=(matrix(1,3,1-sqrt(273),",",6))}}} 

{{{drawing(700,700,-20,20,-18,22,

line(-25,0,25,0), line(0,-25,0,25),line(1,2,-15.62271164,6), line(1,2,17.62271164,6), locate(9,4,17), locate(-8,3.93,17),
locate(1,1.87,"A(1,2)"), red(line(1,2,1,6),locate(1.1,4.5,4)),locate(.4,7,"B(1,6)"),
locate(17.4,7,"C(?,6)"),locate(-18.5,7,"C'(?,6)"),
green(circle(1,2,17),line(-25,6,25,6)) )}}}

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Distance formula:

Let the point (x,y) [where y=6, which means the point (x,6)] have distance
17 from point (1,2).

Then 

{{{d=sqrt((x-1)^2+(6-2)^2)}}}
{{{17=sqrt(x-1)^2+4^2)}}}
{{{17=sqrt((x-1)^2+16)}}}
{{{17=sqrt((x-1)^2+16)}}}
Square both sides:
{{{289=(x-1)^2+16}}}
{{{273=(x-1)^2}}}
{{{"" +- sqrt(273)=x-1}}}
{{{1 +- sqrt(273)=x}}}

Thus the two points are 

{{{(matrix(1,3,1+sqrt(273),",",6))}}} and {{{(matrix(1,3,1-sqrt(273),",",6))}}}

Edwin</pre>