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Sagot :
To determine the slope of the tangent line to circle P at point Q, we need to evaluate several key points about the relationship between the diameter of the circle and the tangent line.
1. Identify the slope of the diameter:
The given equation of the diameter is [tex]\(y = 4x + 2\)[/tex]. This equation is in the slope-intercept form [tex]\(y = mx + c\)[/tex] where [tex]\(m\)[/tex] represents the slope. Therefore, the slope of this line is 4.
2. Understand the relationship between a diameter and a tangent:
A tangent line to a circle at a given point is perpendicular to the radius at that point. Since the radius to point Q lies along the diameter, the tangent at point Q will be perpendicular to the diameter.
3. Find the slope of the tangent line:
If two lines are perpendicular, the product of their slopes is [tex]\(-1\)[/tex]. Denote the slope of the diameter by [tex]\(m_d\)[/tex] and the slope of the tangent by [tex]\(m_t\)[/tex]. Therefore, we have the equation:
[tex]\[ m_d \cdot m_t = -1 \][/tex]
Substituting the known slope of the diameter:
[tex]\[ 4 \cdot m_t = -1 \][/tex]
Solving for [tex]\(m_t\)[/tex]:
[tex]\[ m_t = -\frac{1}{4} \][/tex]
Thus, the slope of the tangent line to circle P at point Q is [tex]\(-0.25\)[/tex].
Considering the options given:
- Option A: Correct. The slope of the tangent line is [tex]\(-0.25\)[/tex].
- Option B: Incorrect. The slope 4 is the slope of the diameter, not the tangent line.
- Option C: Incorrect. The slope -4 doesn't align with our calculation for the tangent line.
- Option D: The string seems incomplete, but it may have intended [tex]\("-0.25"\)[/tex].
Therefore, the correct answer is:
OA. The slope of the tangent line is -0.25.
1. Identify the slope of the diameter:
The given equation of the diameter is [tex]\(y = 4x + 2\)[/tex]. This equation is in the slope-intercept form [tex]\(y = mx + c\)[/tex] where [tex]\(m\)[/tex] represents the slope. Therefore, the slope of this line is 4.
2. Understand the relationship between a diameter and a tangent:
A tangent line to a circle at a given point is perpendicular to the radius at that point. Since the radius to point Q lies along the diameter, the tangent at point Q will be perpendicular to the diameter.
3. Find the slope of the tangent line:
If two lines are perpendicular, the product of their slopes is [tex]\(-1\)[/tex]. Denote the slope of the diameter by [tex]\(m_d\)[/tex] and the slope of the tangent by [tex]\(m_t\)[/tex]. Therefore, we have the equation:
[tex]\[ m_d \cdot m_t = -1 \][/tex]
Substituting the known slope of the diameter:
[tex]\[ 4 \cdot m_t = -1 \][/tex]
Solving for [tex]\(m_t\)[/tex]:
[tex]\[ m_t = -\frac{1}{4} \][/tex]
Thus, the slope of the tangent line to circle P at point Q is [tex]\(-0.25\)[/tex].
Considering the options given:
- Option A: Correct. The slope of the tangent line is [tex]\(-0.25\)[/tex].
- Option B: Incorrect. The slope 4 is the slope of the diameter, not the tangent line.
- Option C: Incorrect. The slope -4 doesn't align with our calculation for the tangent line.
- Option D: The string seems incomplete, but it may have intended [tex]\("-0.25"\)[/tex].
Therefore, the correct answer is:
OA. The slope of the tangent line is -0.25.
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