To determine the slope of a line that is tangent to a circle at a given point [tex]\(Q\)[/tex], we need to understand the relationship between the slope of the diameter and the slope of the tangent line.
Here's a detailed, step-by-step approach to solving the problem:
1. Identify the slope of the given diameter:
The equation of the diameter that passes through point [tex]\(Q\)[/tex] is given as [tex]\(y = 4x + 2\)[/tex]. This equation is in the slope-intercept form [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope of the line. From the equation, we see that the slope [tex]\(m\)[/tex] of the diameter is [tex]\(4\)[/tex].
2. Understand the geometric relationship:
In coordinate geometry, the tangent to a circle at any point is perpendicular to the radius (or a diameter) of the circle at that point. Therefore, the slope of the tangent line at point [tex]\(Q\)[/tex] will be the negative reciprocal of the slope of the diameter.
3. Calculate the negative reciprocal of the slope:
The slope of the diameter is [tex]\(4\)[/tex]. The negative reciprocal of [tex]\(4\)[/tex] is calculated as follows:
[tex]\[
\text{Negative Reciprocal} = -\frac{1}{4}
\][/tex]
4. Determine which statement matches the calculated slope:
The slope of the tangent line at point [tex]\(Q\)[/tex] is [tex]\(-\frac{1}{4}\)[/tex]. This corresponds to option [tex]\(B\)[/tex].
Therefore, the correct answer is:
B. The slope of the tangent line is [tex]\(-\frac{1}{4}\)[/tex].
