Problem 16 - Basic Engineering Mathematics

Question

Sketch the circle given by the equation: $$ x^{2}+y^{2}-4 x+6 y-3=0 $$

Step-by-Step Solution

Answer

Center: \((2, -3)\), Radius: 4.

Step 1: Identify the General Circle Equation

The general equation for a circle is \( (x-h)^2 + (y-k)^2 = r^2 \). Here, \((h, k)\) is the center of the circle, and \(r\) is its radius. The given equation \(x^2 + y^2 - 4x + 6y - 3 = 0\) needs to be rearranged to match this form.

Step 2: Rearrange the Equation

Reorganize the given equation by grouping the \(x\) terms and the \(y\) terms: \((x^2 - 4x) + (y^2 + 6y) = 3\). This will help in completing the square for both \(x\) and \(y\).

Step 3: Complete the Square for x

Take the \(x\) terms \(x^2 - 4x\). Add and subtract \(4\) (\((\frac{-4}{2})^2=4\)): \[(x^2 - 4x + 4 - 4)\]. This becomes \((x-2)^2 - 4\).

Step 4: Complete the Square for y

For the \(y\) terms \(y^2 + 6y\), add and subtract \(9\) (\((\frac{6}{2})^2=9\)): \[(y^2 + 6y + 9 - 9)\]. This simplifies to \((y+3)^2 - 9\).

Step 5: Formulate the Completed Square Equation

Substitute the completed square forms back into the equation: \((x-2)^2 - 4 + (y+3)^2 - 9 = 3\). Simplify to get \((x-2)^2 + (y+3)^2 = 16\). This is the standard circle equation form.

Step 6: Identify the Center and Radius

From the equation \((x-2)^2 + (y+3)^2 = 16\), the center of the circle is \((2, -3)\) and the radius is \(\sqrt{16} = 4\).

Step 7: Sketch the Circle

Draw the coordinate axes. Plot the center of the circle at \((2, -3)\). From the center, measure a distance of \(4\) units in all directions to trace the circle.

Key Concepts

Completing the Square

Completing the square is a technique used to transform a quadratic expression into a perfect square trinomial. This is particularly useful in transforming quadratic equations into their standard forms, such as the circle equation. In our circle example, the original equation is:

• \( x^{2} + y^{2} - 4x + 6y - 3 = 0 \)

The first step involves grouping the terms separately for \(x\) and \(y\) variables, so we have \((x^2 - 4x) + (y^2 + 6y)\). To complete the square for the \(x\) terms, we take the linear coefficient, divide it by 2, and square it. For \(x^2 - 4x\):

• Take \(-4/2 = -2\), square it to get \(4\).
• Add and subtract \(4\) inside the expression to maintain balance.
  This changes our expression to \((x-2)^2 - 4\).

Do the same for \(y\) variables:

• \((y^2 + 6y)\) becomes \((y+3)^2 - 9\).
  Here, \(6/2 = 3\) and \(3^2 = 9\).

Once completed, these squares allow rewriting the circle equation in a standard form.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, allows us to use algebraic methods to solve geometric problems. In the context of a circle, we use coordinate geometry to find the equation of the circle based on its geometric properties, such as its center and radius.
The standard equation of a circle in the coordinate plane is:

• \((x-h)^2 + (y-k)^2 = r^2\)

In this form, \((h,k)\) represents the center of the circle, and \(r\) is the radius. By transforming equations using completing the square, we can express them in this form. This transformation helps us easily identify geometric properties from algebraic equations. For example, by converting our original equation into \((x-2)^2 + (y+3)^2 = 16\), it's evident where the center and radius are located in the coordinate plane. Understanding these conversions bridges algebra and geometry and provides valuable insights into graphical representations of equations.

Radius and Center of a Circle

The radius and center of a circle are fundamental concepts that describe its size and position. The center \((h, k)\) is a fixed point in the plane from which every point on the circle is equidistant, the distance being the radius \(r\).

• The radius is the constant distance from the center \((h, k)\) to any point on the circle.
• In circle equations like \((x-h)^2 + (y-k)^2 = r^2\), transformations help determine these parameters efficiently.

In our example, by completing the square, we obtained:

• \((x-2)^2 + (y+3)^2 = 16\)

Here, the center is \((2, -3)\) and the radius is \(\sqrt{16} = 4\).
This allows us to draw the circle accurately on the coordinate plane.
Knowing how to derive the center and radius not only simplifies graphing tasks but also deepens understanding of the geometric properties of the circle.