Problem 58 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that the set of points that are twice as far from \((3,4)\) as from \((1,1)\) form a circle. Find its center and radius.

Step-by-Step Solution

Answer

The circle's center is (1, 0) and its radius is 2.

Step 1: Define distance equations

Let the coordinates of any point be \((x, y)\). The distance from \((x, y)\) to \((3,4)\) is \(d_1 = \sqrt{(x - 3)^2 + (y - 4)^2}\), and the distance from \((x, y)\) to \((1,1)\) is \(d_2 = \sqrt{(x - 1)^2 + (y - 1)^2}\). According to the problem, \(d_1 = 2d_2\).

Step 2: Set up the equation

Substitute \(d_1\) and \(d_2\) into the condition \(d_1 = 2d_2\) to get: \[ \sqrt{(x - 3)^2 + (y - 4)^2} = 2\sqrt{(x - 1)^2 + (y - 1)^2}. \]

Step 3: Eliminate the square roots

Square both sides of the equation: \[ (x - 3)^2 + (y - 4)^2 = 4((x - 1)^2 + (y - 1)^2). \]

Step 4: Expand both sides

Expand the left side: \((x - 3)^2 + (y - 4)^2 = x^2 - 6x + 9 + y^2 - 8y + 16\). Expand the right side: \(4((x - 1)^2 + (y - 1)^2) = 4(x^2 - 2x + 1 + y^2 - 2y + 1) = 4x^2 - 8x + 4 + 4y^2 - 8y + 4\).

Step 5: Simplify and rearrange

Combine terms and simplify: \(x^2 - 6x + 9 + y^2 - 8y + 16 = 4x^2 - 8x + 4 + 4y^2 - 8y + 4\).Cancel common terms to get: \(-3x^2 + 2x - 3y^2 + 21 = 0\).

Step 6: Contrast the terms

The above simplifies to: \[ x^2 - 2x + y^2 + 3 = 0. \] Rearrange it as: \[ (x - 1)^2 + y^2 = 4. \]

Step 7: Identify center and radius of the circle

The equation of the circle is now in the form \((x - h)^2 + (y - k)^2 = r^2\), with center \((h, k) = (1, 0)\) and radius \(r = 2\).

Key Concepts

Distance Formula

In coordinate geometry, the distance formula is essential for calculating distances between two points on a plane. If you have two points, say \( (x_1, y_1) \) and \( (x_2, y_2) \), you can easily find the distance, \( d \), between them using the formula:

• \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

This formula is derived from the Pythagorean theorem, letting you visualize the distance as the hypotenuse of a right triangle. It's vital when dealing with problems involving circles, especially when determining if points lie on a circle. These calculations form the backbone of coordinate geometry and help solve complex geometric problems with ease.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, allows us to use algebra to investigate geometric properties. It involves the study of geometric figures like lines, curves, and figures using a coordinate system.

• Points are denoted as \( (x, y) \) in a plane, making it easier to analyze distances and relationships.
• Equations like \( (x - 1)^2 + y^2 = 4 \) represent circles, lines, or other shapes based on the coordinates.

In the given problem, coordinate geometry helps us showcase that the set of points maintaining specific distance relations form a circle. It bridges algebra with geometry by expressing spatial figures through equations.

Center of a Circle

The center of a circle is crucial as it defines the circle's position on the plane. In general, a circle can be described by the formula \((x-h)^2 + (y-k)^2 = r^2\), where \( (h, k) \) is the center.

• The position \( (h, k) \) is the reference point from which all points along the circle's edge (radius) are equidistant.

In coordinate geometry, identifying the circle's center allows for understanding its symmetry and spatial orientation. For the problem at hand, identifying \( (1, 0) \) as the center confirms the circular nature of the set of points defined by their distance conditions.

Radius of a Circle

The radius is the constant distance between the center of the circle to any point on its circumference. It appears in the circle's equation \( (x-h)^2 + (y-k)^2 = r^2 \) as \( r^2 \), representing the circle's size.

• In our exercise, you find the radius by completing the equation manipulation, confirming \( r = 2 \).

Understanding the radius helps in visualizing how large the circle is and how it scales on the coordinate plane. It provides a foundation for many applications in geometry and real-world contexts.