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.

Short answer

The circle is centered at \\( (\\frac{1}{3}, 0) \\) with radius \\( \\frac{\\sqrt{26}}{3} \\).

Step-by-step solution

Understand the Distance Formula

To find the distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \), use the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]. We will use this to compare distances from a point \( (x, y) \) to both \((3,4)\) and \((1,1)\).

Set Up the Distance Conditions

Given a point \( (x, y) \), the distance to \( (3, 4) \) is \( \sqrt{(x - 3)^2 + (y - 4)^2} \). The distance to \( (1, 1) \) is \( \sqrt{(x - 1)^2 + (y - 1)^2} \). The problem states the first distance is twice the second, or: \[(x - 3)^2 + (y - 4)^2 = 4((x - 1)^2 + (y - 1)^2)\]

Expand Both Sides of the Equation

First, expand the left-hand side: \[(x - 3)^2 + (y - 4)^2 = x^2 - 6x + 9 + y^2 - 8y + 16\]Combine like terms: \(x^2 + y^2 - 6x - 8y + 25\).Now expand the right-hand side: \[4((x - 1)^2 + (y - 1)^2) = 4(x^2 - 2x + 1 + y^2 - 2y + 1)\]Combine and simplify: \4(x^2 + y^2 - 2x - 2y + 2) = 4x^2 + 4y^2 - 8x - 8y + 8\.

Equate and Simplify the Equation

Set both expanded expressions equal from Step 3:\[ x^2 + y^2 - 6x - 8y + 25 = 4x^2 + 4y^2 - 8x - 8y + 8 \]Bring everything to one side:\[ 0 = 3x^2 + 3y^2 - 2x + 17 \]Simplify by dividing by 3:\[ 0 = x^2 + y^2 - \frac{2}{3}x + \frac{17}{3} \]

Complete the Square for Center and Radius

Rephrase \(-\frac{2}{3}x + \frac{17}{3}\) in terms of a circle equation \( (x-h)^2 + (y-k)^2 = r^2 \).After completing the square for x:\[ (x - \frac{1}{3})^2 = (y - \frac{1}{3})^2 = \frac{26}{9} \]So the circle is centered at \( (\frac{1}{3}, 0) \) with radius \sqrt{\frac{26}{9}} = \frac{\sqrt{26}}{3}\.

Key concepts

Distance Formula

The distance formula is a fundamental concept in geometry used to determine the length between two points in a plane. Consider two points, \((x_1, y_1)\) and \((x_2, y_2)\). The distance \(d\) between these points is calculated using the formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] This formula derives from the Pythagorean theorem, which relates the lengths of the sides of a right triangle. By applying the difference in the x-coordinates and y-coordinates as the two sides of a right triangle, the distance formula finds the hypotenuse of this triangle. When solving problems like identifying circles formed by points equidistant from two other points, the distance formula is essential. It allows you to express the condition that one distance from a point is a specific multiple of another, as in the equation given with the point \((x, y)\).

Circle Equation

The equation of a circle in the Cartesian plane is an essential geometric formula that connects the center of a circle to any point on its circumference. The general form of a circle's equation with center \((h, k)\) and radius \(r\) is given by: \[(x - h)^2 + (y - k)^2 = r^2\] This equation states that all points \((x, y)\) located on the circle are such that their distance from the center \((h, k)\) is constant, specifically the radius \(r\). In problems where you need to verify if a set of points forms a circle, compare them using this equation. For example, showing that the expression \((x - 3)^2 + (y - 4)^2 = 4((x - 1)^2 + (y - 1)^2)\) describes a circle involves transforming it into a standard circle equation through simplifications and reorganizations, as shown in the solution steps.

Completing the Square

Completing the square is a strategic algebraic method used to simplify the format of quadratic equations, making them easier to solve or analyze. When working with circle equations, it often rephrases parts of the equation to fit the canonical form of a circle. For example, to convert a term like \(-\frac{2}{3}x\) appearing in the expanded equation of a circle into a more usable circle format, you can complete the square: - Divide the coefficient of \(x\) by 2, square it, and then add and subtract it within the expression.- This modification rearranges quadratic expressions into perfect squares, such as \((x - \frac{1}{3})^2\). By completing the square for both variable terms in the equation, you convert it to an identifiable circle equation form, from which you can directly read the center and radius of the circle, clarifying the geometrical implications without misinterpretation.

Circle Center and Radius

Identifying the center and radius of a circle from its equation is the final step in analyzing its geometric properties. Once a circle equation is formatted as \((x - h)^2 + (y - k)^2 = r^2\), interpreting the center \((h, k)\) and the radius \(r\) is straightforward. From the completed square form, the circle \((x - \frac{1}{3})^2 + (y - 0)^2 = \left(\frac{\sqrt{26}}{3}\right)^2\) indicates the circle is centered at \(\left(\frac{1}{3}, 0\right)\) with a radius of \(\frac{\sqrt{26}}{3}\). This knowledge is directly useful for graphing the circle, determining symmetry, and understanding its relationship to other geometry problems. Recognizing how to extract these parameters from the equation allows for comprehensive insights into the spatial structure of the circle involved in any geometric analysis.