To solve this question, we need to understand the relationship between the slopes of the given lines. A tangent line to a circle at a point [tex]\( Q \)[/tex] is perpendicular to the line passing through the center of the circle and the point [tex]\( Q \)[/tex]. In this question, the equation of the diameter passing through point [tex]\( Q \)[/tex] is given as [tex]\( y = 4x + 2 \)[/tex].
1. Identify the slope of the given line:
The given line [tex]\( y = 4x + 2 \)[/tex] is in slope-intercept form [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope and [tex]\( b \)[/tex] is the y-intercept.
Here, the slope [tex]\( m \)[/tex] of the given line is [tex]\( 4 \)[/tex].
2. Find the slope of the perpendicular line:
The slope of a line perpendicular to another line is the negative reciprocal of the slope of the original line. So, we need to find the negative reciprocal of [tex]\( 4 \)[/tex].
The negative reciprocal of [tex]\( 4 \)[/tex] is calculated as:
[tex]\[
\text{slope of the perpendicular line} = -\frac{1}{4}
\][/tex]
3. Determine the slope of the tangent line:
Since the tangent line at point [tex]\( Q \)[/tex] on the circle [tex]\( P \)[/tex] is perpendicular to the line passing through the center and the same point [tex]\( Q \)[/tex], the slope of the tangent line must be:
[tex]\[
-\frac{1}{4}
\][/tex]
Therefore, the correct statement describing the slope of the tangent line to circle [tex]\( P \)[/tex] at point [tex]\( Q \)[/tex] is:
C. The slope of the tangent line is [tex]\(-\frac{1}{4}\)[/tex].
