Problem 60 - Calculus with Concepts in Calculus

Question

Show that the points twice as far from the point \((2,-3)\) as from the point \((-1,0)\) form a circle, and find the center and radius of that circle.

Step-by-Step Solution

Answer

The points form a circle centered at (-\frac{7}{3}, \frac{5}{3}) with radius \frac{2\sqrt{5}}{3}.

Step 1: Set up the distance condition

Points \((x,y)\) satisfying: distance to \((2,-3)\) = 2 times distance to \((-1,0)\):
\(\sqrt{(x-2)^2+(y+3)^2} = 2\sqrt{(x+1)^2+y^2}\)

Step 2: Square and simplify

\((x-2)^2+(y+3)^2 = 4[(x+1)^2+y^2]\)
\(x^2-4x+4+y^2+6y+9 = 4x^2+8x+4+4y^2\)
\(3x^2+12x+3y^2-6y-9 = 0\)
\(x^2+4x+y^2-2y-3 = 0\)
\((x+2)^2+(y-1)^2 = 8\)

Step 3: State the result

Center: \((-2, 1)\), Radius: \(2\sqrt{2}\).

Key Concepts

Distance Formula

The Distance Formula is fundamental in circle geometry, as it allows us to calculate the distance between two points in the coordinate plane. The formula can be derived from the Pythagorean Theorem and is expressed as:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]Here,

• \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points.
• The result, \(d\), is the distance between these points.

This formula is essential when trying to maintain specific distance relationships, such as the one described in the exercise, where the distance from \((x, y)\) to \((2, -3)\) is twice the distance from \((x, y)\) to \((-1, 0)\).
The key to solving the exercise involves setting up and solving an equation based on this basic formula to show that the set of points satisfying the condition forms a circle.

Circle Equation

A circle in the coordinate plane can be defined using its equation, which relates directly to its center and radius. The general equation of a circle is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]In this formula,

• \((h, k)\) is the center of the circle.
• \(r\) represents the radius.

In the provided exercise, once the condition using the distance formula was set up, the points were shown to form a circle with a specific center and radius.
By algebraically manipulating the distances, a new equation in a form similar to the general circle equation was derived. This was critical in determining how the points relate to each other and confirming they indeed form a circle.

Geometric Locus

In geometry, a locus is a set of points satisfying a particular condition or rule. The concept of a geometric locus is beautifully illustrated in circle problems. For our current exercise, the locus is described by points whose distance from one fixed point is exactly twice that from another fixed point.

• This specific condition transforms into the equation of a circle.
• Once rewritten using the distance formula, it results in a recognizable circle equation.

In simple terms, the geometric locus here represents the path or location of all points which maintain this relationship — in this case forming a circle. By solving and interpreting the initial distance equation, the exercise showed that such points indeed result in a circle, thus providing a practical application of the concept of a geometric locus.