To find the equation of a line that is tangent to circle [tex]\( P \)[/tex] at point [tex]\( Q \)[/tex], follow these steps:
1. Identify the slope of the diameter:
The equation of the diameter passing through point [tex]\( Q \)[/tex] is given by [tex]\( y = 4x + 2 \)[/tex]. This means the slope of the diameter is 4.
2. Determine the slope of the tangent line:
The tangent to a circle at any point is perpendicular to the radius (diameter) at that point. If two lines are perpendicular, the product of their slopes is -1. Therefore, if the slope of the diameter is 4, the slope of the tangent line, [tex]\( m_t \)[/tex], can be found using:
[tex]\[
4 \cdot m_t = -1
\][/tex]
Solving for [tex]\( m_t \)[/tex]:
[tex]\[
m_t = -\frac{1}{4}
\][/tex]
3. Compare with the given choices:
- A. The slope of the tangent line is [tex]\(\frac{1}{4}\)[/tex]. This is incorrect as our result shows the slope should be [tex]\(-\frac{1}{4}\)[/tex].
- B. The slope of the tangent line is -4. This is also incorrect because -4 is not the negative reciprocal of 4.
- C. The slope of the tangent line is 4. This is incorrect as 4 is the slope of the diameter, not the tangent.
- D. The slope of the tangent line is [tex]\(-\frac{1}{4}\)[/tex]. This is correct according to our calculation.
Therefore, the correct answer is:
D. The slope of the tangent line is [tex]\(-\frac{1}{4}\)[/tex].
