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Let k be a positive real number. The line x + y = 3k and the circle x^2 + y^2 = k - 4x - 6y are drawn. Find $k$ so that the line is tangent to the circle.

 May 10, 2024
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y  = -x + 3k  ......the slope of this line   = -1

 

x^2 + 4x + y^2 + 6y =  k      complete the square on x,y

 

x^2 + 4x + 4 + y^2 + 6y + 9  =  k + 4 + 9

 

(x + 2)^2  + (y + 3)^2  =  13 + k

 

The center of this  circle is  (-2, -3)

 

A line with a slope = 1  through this point will intersect the tangent line

y = (x +2) - 3

y = x -1

 

So

x -1 = -x + 3k

2x = 3k+1

x = (3k + 1) / 2

 

y = (3k+ 1)  / 2  - 1

y = (3k + 1 - 2) / 2

y = (3k - 1) / 2

 

Putting these values into the equation of the circle to find k

 

[ (3k + 1) / 2 + 2]^2   +[ (3k -1)/2 + 3[^2  =  13 + k

 

Simplifying this we  get

 

9k^2  + 28k - 1  =  0

 

Use the Quad Formula

 

We have two values for k

k ≈ .035313

and

k ≈ -3.1464

 

cool cool cool

 May 10, 2024

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