UpStudy Free Solution:

This can be done by substituting the equation of the line into the equation of the circle and checking whether the resulting quadratic equation has real solutions.
First, substitute \(y = 2x - 1\) into the equation of the circle:
\[x^ 2 + ( 2x - 1) ^ 2 + 15x - 11( 2x - 1) + 9 = 0\]
Now, simplify the equation:
\[x^ 2 + ( 4x^ 2 - 4x + 1) + 15x - 22x + 11 + 9 = 0\]
Combine like terms:
\[x^ 2 + 4x^ 2 - 4x + 1 + 15x - 22x + 11 + 9 = 0\]
\[5x^ 2 - 11x + 21 = 0\]
We have a quadratic equation in the form \(ax^ 2 + bx + c = 0\), where \(a = 5\), \(b = - 11\), and \(c = 21\). To determine whether this quadratic equation has real solutions, we can use the discriminant \(\Delta \):
\[\Delta = b^ 2 - 4ac\]
Substitute the values of \(a\), \(b\), and \(c\):
\[\Delta = ( - 11) ^ 2 - 4( 5) ( 21) \]
\[\Delta = 121 - 420\]
\[\Delta = - 299\]
Since the discriminant \(\Delta \) is negative (\(\Delta = - 299\)), the quadratic equation \(5x^ 2 - 11x + 21 = 0\) has no real solutions. This means that the line \(y = 2x - 1\) and the circle \(x^ 2 + y^ 2 + 15x - 11y + 9 = 0\) do not intersect.
Therefore, we have shown that the line and the circle do not intersect.
 

Supplemental Knowledge

In analytic geometry, the intersection of a line and a circle can be determined by substituting the equation of the line into the equation of the circle and solving for the points of intersection. If the resulting quadratic equation has no real solutions, it means that the line does not intersect the circle.
The general form of a circle's equation is:
\[x^ 2 + y^ 2 + Dx + Ey + F = 0\]
where \(D\), \(E\), and \(F\) are constants. The general form of a line's equation is:
\[y = mx + c\]
where \(m\) is the slope and \(c\) is the y-intercept.

UpStudy geometry offers precise solutions with step-by-step explanations tailored specifically to you, plus elite tutors available 24/7 online as needed.