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Chapter 2: Problem 52

Suppose that a circle is tangent to both axes, is in the third quadrant, and has radius \(\sqrt{2} .\) Find the center-radius form of its equation.

Short Answer

Expert verified
(x + \sqrt{2})^2 + (y + \sqrt{2})^2 = 2.

Step by step solution

01

Understand the problem

Identify the characteristics of the circle being tangent to both axes and locate the third quadrant. The circle will have a radius of \(\sqrt{2} \) and will touch both the negative x-axis and y-axis.
02

Determine the center coordinates

Since the circle is tangent to both axes and in the third quadrant, the center of the circle will be \((-a, -a) \) where \(\ a \). Since it has to be tangent to both axes, the distance from the center to each axis equals the radius \(\sqrt{2} \). Therefore, the center of the circle is at \((-\sqrt{2},-\sqrt{2} \).
03

Write the center-radius form

The center-radius form of a circle's equation is \((x-h)^2 + (y-k)^2 = r^2 \) where \((h, k) \) is the center and \(\ r \). Using the center \((-\sqrt{2}, -\sqrt{2}) \) and radius \(\sqrt{2} \), substitute these values into the equation. This results in \((x + \sqrt{2})^2 + (y + \sqrt{2})^2 = 2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

center-radius form
Understanding the center-radius form of a circle is crucial in working with circle equations. The center-radius form is written as \( (x-h)^2 + (y-k)^2 = r^2 \), where \( (h, k) \) are the coordinates of the circle's center and \( r \) is the radius.
In this form:
  • \( (x-h) \) represents the horizontal distance from any point on the circle to the center.
  • \( (y-k) \) represents the vertical distance from any point on the circle to the center.
  • \( r^2 \) is the square of the radius of the circle.

When you know the center and radius, you can simply plug these into the equation. For instance, for a circle with center \((-h, -k)\) and radius \( r \), the equation is \((x+h)^2 + (y+k)^2 = r^2\). This equation helps visualize and solve problems involving circles efficiently.
tangent to axes
A circle is said to be tangent to an axis if it touches the axis at exactly one point. For a circle tangent to both axes:
  • The distance from the circle鈥檚 center to each axis equals the radius.
  • If the circle is in the third quadrant, its center will have negative coordinates.

In the given problem, since the circle is tangent to both the x-axis and y-axis, and is in the third quadrant, its center must be \((-a, -a)\). We know this because:
  • The circle touches both negative axes, indicating the center is equally distant from both.
  • The radius extends from the center to both axes, indicating \(a=r \).

For our circle with radius \(\rad2\), the center thus becomes \((-\rad2, -\rad2)\).
third quadrant
The coordinate plane is divided into four quadrants, each representing a distinct region based on the signs of the x and y coordinates. The third quadrant is crucial for understanding circle placements in problems:
  • The third quadrant encompasses points where both x and y coordinates are negative.
  • This is why a circle in this quadrant with a center \((h, k)\) will have both h and k as negative values.

In this example, our circle, with a radius of \(\rad2\), has its center at \((-\rad2, -\rad2)\). This fits the third quadrant's rule that both coordinates are negative and ensures the circle's edges just touch both axes at the points \((0, -\rad2) \) and \((-\rad2, 0)\).
Understanding the specific quadrant helps determine positioning and distances accurately in geometrical problems.

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Most popular questions from this chapter

Let \(f(x)=2 x-3\) and \(g(x)=-x+3 .\) Find each function value. $$(f \circ g)(4)$$

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