To determine the slope of the tangent line to circle [tex]\( P \)[/tex] at point [tex]\( Q \)[/tex], given that the slope of the diameter through point [tex]\( Q \)[/tex] is described by the equation [tex]\( y = 4x + 2 \)[/tex], we need to follow these steps:
1. Identify the slope of the diameter line:
The given equation of the line passing through the diameter is [tex]\( y = 4x + 2 \)[/tex]. This equation is in the slope-intercept form [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] represents the slope. Thus, the slope of the diameter line is 4.
2. Determine the slope of the tangent line:
The tangent line to a circle at any point is perpendicular to the radius (or diameter) at that point. Therefore, the slope of the tangent line is the negative reciprocal of the slope of the diameter line.
To find the negative reciprocal of the slope [tex]\( 4 \)[/tex]:
- The reciprocal of [tex]\( 4 \)[/tex] is [tex]\( \frac{1}{4} \)[/tex].
- The negative reciprocal of [tex]\( 4 \)[/tex] is [tex]\( -\frac{1}{4} \)[/tex].
3. Select the correct statement:
The slope of the tangent line is [tex]\( -\frac{1}{4} \)[/tex].
Therefore, the answer is:
A. The slope of the tangent line is [tex]\( -\frac{1}{4} \)[/tex].
