Math Problem Statement
Find the initial point on the part of circle 𝑥2 + 𝑦2 = 16 in 3rd quadrant
Solution
The equation of the circle is:
This is the equation of a circle with a radius of 4 (since ) and centered at the origin .
To find the initial point on the circle in the third quadrant, we need to focus on the coordinates that satisfy both:
- The equation of the circle.
- The conditions for being in the third quadrant, where both and .
Step 1: Parameterize the equation
The general parametric form of a point on the circle is:
Step 2: Identify the angle in the third quadrant
In the third quadrant, both and are negative, so we need to find an angle such that:
This condition is satisfied when lies between and , or equivalently between and radians.
Step 3: Choose a specific point
Let's consider (the angle where the point is at the far left of the circle, on the negative -axis). For this angle:
This is a point on the -axis (not in the third quadrant). To get a point specifically in the third quadrant, we choose a value like , which is in the third quadrant.
For :
Thus, the initial point on the circle in the third quadrant is approximately .
Would you like further clarification on this solution? Here are some additional questions related to this topic:
- What is the general formula for parametric equations of a circle?
- How can you find the points in the other quadrants for the same circle?
- What does it mean for a point to be in the third quadrant?
- How can you derive the parametric equations for a circle from its standard equation?
- What happens to the coordinates of a point on the circle as you move through different quadrants?
Tip: In parametric equations, using specific angles (like , , or ) helps in locating distinct points on a circle.
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Math Problem Analysis
Mathematical Concepts
Circle Geometry
Parametric Equations
Trigonometry
Formulas
x = r * cos(θ)
y = r * sin(θ)
x² + y² = r²
Theorems
Properties of Circles
Parametric Equation of a Circle
Suitable Grade Level
Grades 10-12
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