Question

Find all values of k so that the following system has no solutions x²+(y-k)²=36 ; y=5x+k.Illustrate with a graph.

Answer 1

Step-by-step explanation:

$x^2+(y-k)^2=36$

$y=5x+k$

Substituting y in the first equation

$x^2+(5x+k-k)^2=36\\\Rightarrow x^2+25x^2=36\\\Rightarrow 26x^2=36\\\Rightarrow x^2=\frac{36}{26}\\\Rightarrow x=\sqrt{\frac{36}{26}}\\\Rightarrow x=\pm 1.17$

At x = 1.17

$y=5\times 1.17+k\\\Rightarrow y=5.85+k$

At x = -1.17

$y=5\times -1.17+k\\\Rightarrow y=-5.85+k$

The intersection points are (1.17, 5.85k) and (-1.17, -5.85k)

The graph would intersect regardless of what value k is.