To determine the slope of the tangent line to circle [tex]\(P\)[/tex] at point [tex]\(Q\)[/tex], start with the information that the equation of the diameter passing through point [tex]\(Q\)[/tex] is [tex]\(y = 4x + 2\)[/tex].
1. Identify the Slope of the Diameter:
- The equation [tex]\(y = 4x + 2\)[/tex] is in slope-intercept form ([tex]\(y = mx + b\)[/tex]), where [tex]\(m\)[/tex] is the slope.
- The slope of the diameter is thus [tex]\(4\)[/tex].
2. Tangent Line Relationship:
- A tangent line at any point on a circle is perpendicular to the radius (or diameter) at that point.
- If two lines are perpendicular, the product of their slopes is [tex]\(-1\)[/tex].
3. Find the Slope of the Tangent Line:
- Let [tex]\(m_{\text{diameter}}\)[/tex] be the slope of the diameter, which is [tex]\(4\)[/tex].
- The slope [tex]\(m_{\text{tangent}}\)[/tex] of the tangent line will be the negative reciprocal of [tex]\(m_{\text{diameter}}\)[/tex].
- [tex]\(m_{\text{tangent}} = -\frac{1}{m_{\text{diameter}}}\)[/tex].
- Therefore, [tex]\(m_{\text{tangent}} = -\frac{1}{4}\)[/tex].
4. Conclusion:
- The slope of the tangent line is [tex]\(-\frac{1}{4}\)[/tex].
Thus, the correct answer is:
D. The slope of the tangent line is [tex]\(-\frac{1}{4}\)[/tex].
