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:

x2+y2=16x^2 + y^2 = 16

This is the equation of a circle with a radius of 4 (since 16=4\sqrt{16} = 4) and centered at the origin (0,0)(0, 0).

To find the initial point on the circle in the third quadrant, we need to focus on the coordinates that satisfy both:

  1. The equation of the circle.
  2. The conditions for being in the third quadrant, where both x<0x < 0 and y<0y < 0.

Step 1: Parameterize the equation

The general parametric form of a point on the circle is:

x=4cos(θ)x = 4 \cos(\theta) y=4sin(θ)y = 4 \sin(\theta)

Step 2: Identify the angle in the third quadrant

In the third quadrant, both xx and yy are negative, so we need to find an angle θ\theta such that:

cos(θ)<0andsin(θ)<0\cos(\theta) < 0 \quad \text{and} \quad \sin(\theta) < 0

This condition is satisfied when θ\theta lies between 180180^\circ and 270270^\circ, or equivalently between π\pi and 3π2\frac{3\pi}{2} radians.

Step 3: Choose a specific point

Let's consider θ=π\theta = \pi (the angle where the point is at the far left of the circle, on the negative xx-axis). For this angle:

x=4cos(π)=4×(1)=4x = 4 \cos(\pi) = 4 \times (-1) = -4 y=4sin(π)=4×0=0y = 4 \sin(\pi) = 4 \times 0 = 0

This is a point on the xx-axis (not in the third quadrant). To get a point specifically in the third quadrant, we choose a value like θ=5π4\theta = \frac{5\pi}{4}, which is in the third quadrant.

For θ=5π4\theta = \frac{5\pi}{4}:

x=4cos(5π4)=4×(12)=4×12=22x = 4 \cos\left(\frac{5\pi}{4}\right) = 4 \times \left(-\frac{1}{\sqrt{2}}\right) = -4 \times \frac{1}{\sqrt{2}} = -2\sqrt{2} y=4sin(5π4)=4×(12)=4×12=22y = 4 \sin\left(\frac{5\pi}{4}\right) = 4 \times \left(-\frac{1}{\sqrt{2}}\right) = -4 \times \frac{1}{\sqrt{2}} = -2\sqrt{2}

Thus, the initial point on the circle in the third quadrant is approximately (22,22)(-2\sqrt{2}, -2\sqrt{2}).


Would you like further clarification on this solution? Here are some additional questions related to this topic:

  1. What is the general formula for parametric equations of a circle?
  2. How can you find the points in the other quadrants for the same circle?
  3. What does it mean for a point to be in the third quadrant?
  4. How can you derive the parametric equations for a circle from its standard equation?
  5. 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 π/4\pi/4, π/2\pi/2, or 3π/43\pi/4) 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