Problem 6 - Algebra 2 and Trigonometry

Question

In \(3-10\) , the coordinates of point \(P\) on the circle with center at \(C\) are given. Write an equation of each circle: a. in center-radius form b. in standard form. $$ P(3,2), C(4,2) $$

Step-by-Step Solution

Answer

Center-radius form: \((x-4)^2 + (y-2)^2 = 1\); Standard form: \(x^2 + y^2 - 8x - 4y + 19 = 0\).

Step 1 - Identify the components needed for the circle equation

We know the center of the circle \(C\) is at \((4,2)\) and a point \(P\) on the circle is at \((3,2)\). The center-radius form requires the center \((h,k)\) and the radius \(r\).

Step 2 - Calculate the radius

The radius \(r\) is the distance between the center \(C(4,2)\) and the point \(P(3,2)\). Use the distance formula: \( r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \). Substituting the values gives \( r = \sqrt{(3-4)^2 + (2-2)^2} = \sqrt{1^2 + 0^2} = 1 \).

Step 3 - Write the equation in center-radius form

The center-radius form of the equation for a circle is \((x-h)^2 + (y-k)^2 = r^2\). Substituting \(h = 4\), \(k = 2\), and \(r = 1\), the equation becomes \((x-4)^2 + (y-2)^2 = 1^2\) or \((x-4)^2 + (y-2)^2 = 1\).

Step 4 - Expand to standard form

The standard form of a circle's equation is \(x^2 + y^2 + Dx + Ey + F = 0\). Expanding \((x-4)^2 + (y-2)^2 = 1\) gives: \((x^2 - 8x + 16) + (y^2 - 4y + 4) = 1\). Simplifying, we have \(x^2 + y^2 - 8x - 4y + 19 = 0\) as the standard form.

Key Concepts

Center-Radius Form

When we talk about the center-radius form of a circle equation, it's all about simplifying everything based on the circle's center and its radius. This form is very straightforward to understand and use. The formula looks like this:\[(x-h)^2 + (y-k)^2 = r^2\]- Here, - "\((h, k)\)" represents the coordinates of the circle's **center**. - "\(r\)" is the **radius** of the circle, which is the distance from the center to any point on the circle.So for the given circle, with a center at \((4, 2)\) and the point \((3, 2)\) on it, we first find the radius using the distance formula (we'll discuss this next!). After we have our radius, we can plug those values into the center-radius equation. For this exercise, the center-radius equation is:\[(x-4)^2 + (y-2)^2 = 1\]This form uses the circle's direct information, making it clear what the center and radius are at a glance.

Standard Form of a Circle

Now let's move to the **standard form** of a circle's equation. This form provides a different way to express the same circle and is particularly useful for algebraic manipulation and integration into other mathematical contexts. The general appearance in standard form is:\[x^2 + y^2 + Dx + Ey + F = 0\]- This form allows visibility and synthesis with other linear and geometric equations, providing flexibility in multi-equation environments.- By expanding the center-radius form, you arrive at the standard form, which simplifies the terms and reveals interesting properties of the circle within the plane.For our exercise, after expanding the center-radius form \((x-4)^2 + (y-2)^2 = 1\), we get:\[x^2 + y^2 - 8x - 4y + 19 = 0\]In this format, all terms are balanced with respect to the plane's axes, making it less direct but integral to comprehensive equation systems.

Distance Formula

The **distance formula** is fundamental in calculating how far apart two points are in the coordinate plane. This is especially handy in geometry when working with shapes like circles. The distance formula is expressed as follows:\[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]- In our context, this formula helps us determine the radius of the circle, given the center and a point on the circle's edge.- It's derived from the Pythagorean theorem, embodying how coordinates translate into spatial distances visually and mathematically.For our specific example where - the center of the circle is \(C(4,2)\), and - a point on the circle is \(P(3,2)\), we plug these coordinates into the formula to get:\[r = \sqrt{(3-4)^2 + (2-2)^2} = \sqrt{1^2 + 0^2} = 1\]The result, 1, indicates the radius. Using the distance formula allows the effective transition from known points to numerical radius, grounding many concepts into tangible space.