Find all values of k such that the circle whose equation is x^2 + y^2 + 8x - 12y + k^2 = 0 is tangent to the x-axis.

Question

this was for our homework and i have no idea how to do it :(((

Answer 1

x² + y² + 8x - 12y = -k²

 

Complete the square for the x and y quadratic forms, respectively:

(x² + 8x + 16) + (y² - 12y + 36) = -k² + 16 + 36

 

(x+4)² + (y-6)² = 52-k²

 

Compare this to the standard form of a circle (x-c_x)² + (y-c_y)² = r²  where (c_x,c_y) = center and r=radius,

and solve for k such that r is just big enough for the circle to touch the x-axis.  

 

I get k=±4  

 

 

Answer 2

Complete the square:

 

    (x²+8x+16) + (y²-12y+36) = 52-k²

 

    (x+4)² + (y-6)² = 52-k²

 

       The center is (-4, 6).  The circle will be tangent to the x-axis if the radius is 6.

 

       So, we want to choose k so that 52-k² = 6² = 36

 

                                                          -k²= -16

 

                                                           k = ±4