To find the equation of a line that is tangent to circle [tex]\( P \)[/tex] at point [tex]\( Q \)[/tex], we first need to understand the relationship between the diameter of the circle and the tangent line at that point.
1. Determine the Slope of the Diameter:
Given the equation of the diameter that passes through point [tex]\( Q \)[/tex]:
[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. From the given equation, the slope [tex]\( m \)[/tex] (slope of the diameter) is:
[tex]\[
m = 4
\][/tex]
2. Find the Slope of the Tangent Line:
The slope of a line tangent to a circle at a given point is the negative reciprocal of the slope of the diameter at that same point. Hence, we take the negative reciprocal of 4:
[tex]\[
\text{slope of the tangent line} = -\frac{1}{4}
\][/tex]
Given the choices, the statement that best describes the equation of a line that is tangent to circle [tex]\( P \)[/tex] at point [tex]\( Q \)[/tex] is:
[tex]\[
\boxed{\text{A. The slope of the tangent line is } -\frac{1}{4}.}
\][/tex]
