Question 1048633
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Determine the value(s) of k such that the circle x^2+(y-6)^2 = 36  and the parabola x^2 = 4ky  will intersect only at the origin.
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<pre>
{{{x^2+(y-6)^2}}} = 36,   (1)
{{{x^2}}} = 4ky           (2)


The circle {{{x^2+(y-6)^2}}} = {{{36}}} has the center at (x,y) = (0,6) and has the radius of 6. 
So, the circle has the y-axis x=0 as a diameter and as a symmetry line, passes through the origin and touches the x-axis.

Parabola {{{x^2}}} = {{{4ky}}} also passes through the origin; has the y-axis x=0 as its symmetry line, and touches the x-axis.

After these geometric considerations (that are useful but are not absolutely necessary) let solve the problem algebraically.

Based on (2), substitute 4ky instead of {{{x^2}}} into the equation (1). You will get

{{{4ky + (y-6)^2}}} = 36.

So, in this way you excluded "x" from the system and got a single equation for "y". Let us simplify it:

{{{4ky + y^2 - 12y + 36}}} = 36,  or

{{{y^2 + (4k-12)*y}}} = 0.  (1)

Now, the problem requires this equation (1) to have only one non-negative solution.
     (One solution is evident/obvious. It is y = 0.)

It implies that (4k-12) MUST be non-positive:  4k-12 <= 0.
OTHERWISE y = {{{-(4k-12)}}} would be the other non-negative solution to (1).

So, the solution to the problem is this inequality 4k-12 <= 0,  or, equivalently,  k <= 3   (3 = {{{12/4}}})


<U>Answer</U>.  k <= 3.
</pre>

See an illustration below for k = 3, 2, and 4.


{{{graph( 330, 330, -9, 9, -2.5, 15.5,
          6 + sqrt(36-x^2), 6-sqrt(36-x^2), x^2/12, x^2/8, x^2/16
)}}}


The circle {{{x^2+(y-6)^2}}} = {{{36}}} (red + green) 
and three parabolas x^2 = 4ky for k=3 (blue), k=2, and k=4. 



{{{graph( 330, 330, -4.5, 4.5, -1.5, 7.5,
          6 + sqrt(36-x^2), 6-sqrt(36-x^2), x^2/12, x^2/8, x^2/16
)}}}


The circle {{{x^2+(y-6)^2}}} = {{{36}}} (green) 
and three parabolas x^2 = 4ky for k=3 (blue), k=2, and k=4. 



For solution of similar problems see the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Systems-of-equations/Solving-the-system-of-algebraic-equations-of-degree-2.lesson>Solving systems of algebraic equations of degree 2</A> 

in this site.


Also, you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.