Problem 44 - Precalculus Essentials

Question

Write the standard form of the equation of the circle with the given center and radius. $$(x-2)^{2}+(y-3)^{2}=16$$

Step-by-Step Solution

Answer

The standard form of the equation of the circle is \(x^{2}+y^{2}-4x-6y-3=0\).

Step 1: Understand the given equation

The given equation of the circle is \((x-2)^{2}+(y-3)^{2}=16\). This form is readable as \((x-h)^{2}+(y-k)^{2}=r^{2}\), where (h,k) represents the center and r represents the radius. In this case, the center is (2,3) and the radius is \(\sqrt{16}\) which simplifies to 4.

Step 2: Rewrite the equation

The standard form of the equation of a circle is \(x^{2}+y^{2}+Dx+Ey+F=0\). To write the given equation in standard form, you need to expand and rearrange it.

Step 3: Expand the squares

Expand \((x-2)^{2}\) and \((y-3)^{2}\) to get \(x^{2}-4x+4+y^{2}-6y+9=16\).

Step 4: Combine like terms

Combine like terms to obtain the standard form: \(x^{2}+y^{2}-4x-6y+4+9-16=0\).

Step 5: Simplify the equation

Once you simplify the equation, you get \(x^{2}+y^{2}-4x-6y-3=0\), which is the equation of the circle in its standard form.

Key Concepts

Standard Form of a Circle Equation

The standard form of a circle equation is essential in understanding the layout of a circle on a coordinate plane. It appears as \[x^2 + y^2 + Dx + Ey + F = 0\],where

• \(D\), \(E\), and \(F\) are constants.
• This form gives a quadratic equation when plotted.

To transform a circle's equation from another form into this standard form, as seen in the exercise, involves expanding expressions and rearranging terms to fit this format.

The original exercise starts with a circle equation in a recognizable pattern: \((x-2)^2 + (y-3)^2 = 16\). The task was to express it in standard form by expanding the binomials first, and then combining like terms, resulting in: \[x^{2}+y^{2}-4x-6y-3=0.\] This form makes it easier to analyze and graph the circle, highlighting its geometric properties neatly.

Center-Radius Form of a Circle Equation

The center-radius form of a circle equation is often the most intuitive. It looks like: \[(x-h)^2 + (y-k)^2 = r^2\], where:

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

This version directly shows the circle's center and size, which helps in visualizing the circle right on the plane.

From the original equation given, \((x-2)^{2}+(y-3)^{2}=16\), we learn that the circle's center is at \((2, 3)\) with a radius of \(4\), because \(r = \sqrt{16}\).

When dealing with circle problems, recognizing or converting to center-radius form helps you quickly determine visualization aspects, dealing directly with the geometric shape.

Expanding Binomials

Expanding binomials is a crucial algebraic skill you need when handling circle equations. In this context, it's used to help translate a circle's equation from center-radius form to standard form.

To expand a binomial, recall that \((a-b)^2\) can be rewritten as:

• \(a^2 - 2ab + b^2\).

This principle was applied when expanding the parts of the given circle equation \((x-2)^2 + (y-3)^2 = 16\).

Firstly, \((x-2)^2\) expands to \(x^2 - 4x + 4\).
Similarly, \((y-3)^2\) becomes \(y^2 - 6y + 9\).

Combining all expanded parts, we obtain \[x^2 + y^2 - 4x - 6y + 4 + 9 = 16.\]The sum of like terms and rearranging leads us nicely into solving for the circle equation in standard form. This method demonstrates that understanding the property of binomials simplifies complex algebraic problems.