Problem 56 - Precalculus

Question

Find the standard equation of the circle passing through the origin and with center (3,5)

Step-by-Step Solution

Answer

The circle's equation is \((x-3)^2 + (y-5)^2 = 34\).

Step 1: Identifying the Known Parameters

We know that the center of the circle is given as (3, 5) and the circle passes through the origin (0,0). From this information, we can extract the center coordinates as \((h, k) = (3, 5)\).

Step 2: Calculating the Radius

The radius of the circle is the distance from the center (3, 5) to the point on the circle it passes through, which is the origin (0, 0). Use the distance formula \( r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) to find \[ r = \sqrt{(3-0)^2 + (5-0)^2} \] which simplifies to \[ r = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} \].

Step 3: Writing the Standard Equation

The standard form of the equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\). Substitute the values of \((h, k) = (3, 5)\) and \(r = \sqrt{34}\) into the equation:\[(x - 3)^2 + (y - 5)^2 = (\sqrt{34})^2\]The equation of the circle becomes \((x-3)^2 + (y-5)^2 = 34\).

Key Concepts

Center of a Circle

The center of a circle is a crucial concept in geometry and plays an important role in forming a circle's equation. It is represented by a set of coordinates

• The center provides a reference point that determines the circle's location in the Cartesian plane. This information allows us to describe the positional characteristics of a circle with ease, simplifying geometric and algebraic calculations.
• For any given circle, the center is often denoted as \((h, k)\), where \(h\) and \(k\) are the x and y coordinates, respectively.
• In the given exercise, the center is located at \((3, 5)\), placing it three units along the positive x-axis and five units along the positive y-axis.

By knowing the center of a circle, we can easily relate it to other geometric elements of the shape, such as its radius and circumference. This also provides valuable insight when graphing the circle or performing transformations in coordinate geometry.

Radius Calculation

The radius of a circle is the distance from its center to any point on its circumference. This distance is consistent for all circles, making radius calculation straightforward but significant in forming the circle's equation.

• To find the radius, we often use the distance formula, especially when given two points: the center and a point on the circumference.
• In our exercise, these points are the center \((3, 5)\) and the origin \((0, 0)\), where the circle passes through.

It's essential to remember the distance formula:

• \( r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
• where \((x_1, y_1)\) and \((x_2, y_2)\) are coordinates of the two points.

Applying the formula for our circle gives us \( r = \sqrt{34} \). Using this calculated radius, we can then easily write the full equation of the circle.

Distance Formula

The distance formula is an essential tool in geometry that helps us measure the length between two points in a coordinate plane. It’s a practical application of the Pythagorean theorem.

• The formula calculates the absolute distance, ensuring the value is non-negative, which is crucial for finding components like the radius of a circle.
• For any two points \((x_1, y_1)\) and \((x_2, y_2)\), the distance formula is:\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \].

This wisdom can be leveraged in various geometric applications, not just circle problems, to identify distances between points with precision.In our problem, we use this formula to compute the radius of a circle by treating the center and an edge point as coordinate pairs. This calculation delivers a pivotal measurement for describing the circle fully in mathematical terms, such as when writing its equation in standard form as

• \((x - h)^2 + (y - k)^2 = r^2\).

This precise form allows mathematicians and students alike to graph circles or apply further transformations and alterations with confident accuracy.