To find the slope of the tangent line to the circle [tex]\( P \)[/tex] at point [tex]\( Q \)[/tex], we need to use the fact that the tangent line is perpendicular to the radius (or diameter) at the point of tangency. Given that the diameter of the circle passing through [tex]\( Q \)[/tex] has the equation [tex]\( y = 4x + 2 \)[/tex], we can determine the slope of the diameter and subsequently the slope of the tangent line.
1. Identify the slope of the diameter:
The given equation of the diameter line is [tex]\( y = 4x + 2 \)[/tex]. The general form of a linear equation is [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] represents the slope. Comparing this general form to our given equation, we see that the slope of the diameter ([tex]\( m \)[/tex]) is 4.
2. Determine the slope of the tangent line:
A fundamental property of perpendicular lines in analytic geometry is that the product of their slopes is [tex]\(-1\)[/tex]. Let's denote the slope of the tangent line as [tex]\( m_t \)[/tex]. Then, we use the condition of perpendicularity:
[tex]\[
\text{slope of diameter} \times \text{slope of tangent line} = -1
\][/tex]
Substituting the known slope of the diameter into this equation gives:
[tex]\[
4 \times m_t = -1
\][/tex]
3. Solve for the slope of the tangent line:
[tex]\[
m_t = \frac{-1}{4}
\][/tex]
Thus, the slope of the tangent line to circle [tex]\( P \)[/tex] at point [tex]\( Q \)[/tex] is [tex]\(-\frac{1}{4}\)[/tex].
So, the correct statement is:
D. The slope of the tangent line is [tex]\(-\frac{1}{4}\)[/tex].
