Question 1184285
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If two circles {{{ x^2 + y^2 + 2x + 2ky + 6 = 0 }}} and {{{  x^2 + y^2 + 2ky + k = 0 }}} intersect orthogonally, then what is the value of k?
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            As posed,  this problem  HAS  NO  SOLUTION,


            and I will show it  RIGHT  NOW.



<pre>
First circle

    {{{x^2 + y^2 + 2x + 2ky + 6 = 0}}}

    {{{(x^2 + 2x)}}} + {{{y^2 + 2ky)}}} = -6

    {{{(x+1)^2 - 1}}} + {{{(y+k)^2 - k^2}}} = -6

    {{{(x+1)^2}}} + {{{(y+k)^2}}} = - 6 + 1 + k^2

    {{{(x+1)^2}}} + {{{(y+k)^2}}} = k^2 - 5

has the center at the point (-1,-k)  and the radius of  {{{sqrt(k^2-5)}}}.

Note that the condition  k^2 >= 5 is the NECESSARY condition for existing such a circle.



Second circle 

    {{{x^2 + y^2 + 2ky + k = 0}}}

    {{{x^2}}} + {{{y^2 + 2ky}}} = -k

    {{{x^2}}} + {{{(y+k)^2 - k^2}}} = -k

    {{{x^2}}} + {{{(y+k)^2}}} = k^2-k

has the center at the point (0,-k)  and the radius of  {{{sqrt(k^2-k)}}}.  

Note that the condition  k^2 - k >= 0 is the NECESSARY condition for existing such a circle.



The line connecting the centers is horizontal y = -k,  and the distance between the centers is 1 unit.


So, we have a right angled triangle, formed by the line of centers as the hypotenuse of 1 unit long,

and the legs, that are the radii, drawn to the intersection point, of the lengths found above.


Next, we write the Pythagorean equation

    {{{(k^2-5)}}} + {{{(k^2-k)}}} = 1,


which gives
   
    2k^2 - k - 6 = 0.


From this equation, the roots are

    {{{k[1,2]}}} = {{{(1 +- sqrt(1 + 4*2*6))/(2*2)}}} = {{{(1 +- sqrt(49))/4}}}.


Thus the two possible roots are k= 2 and k= -1.5.      


But no one of these values of  k  satisfies the necessary condition  k^2 >= 5.


THEREFORE, the problem, as it is posed, HAS NO SOLUTION.
</pre>


Solved &nbsp;&nbsp;(or better to say, &nbsp;DISPROVED).


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For your safety, &nbsp;IGNORE &nbsp;the post by @MathLover1, &nbsp;since her answer is &nbsp;WRONG.