Problem 70 - Calculus Single Variable

Question

Find the centers of the two circles of radius \(\sqrt{65}\) that pass through the points (0,-6) and (3,-5) .

Step-by-Step Solution

Answer

The centers are (-3, 0) and (4, -11).

Step 1: Understanding the Problem

We need to find the centers of two circles with radius \(\sqrt{65}\) that pass through the points (0, -6) and (3, -5).

Step 2: Setting up the General Equation of a Circle

The general equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. Here, \(r = \sqrt{65}\).

Step 3: Using Point (0, -6)

Substitute \((x, y) = (0, -6)\) into the circle equation: \((0 - h)^2 + (-6 - k)^2 = (\sqrt{65})^2\), which simplifies to \(h^2 + (k + 6)^2 = 65\).

Step 4: Using Point (3, -5)

Substitute \((x, y) = (3, -5)\) into the circle equation: \((3 - h)^2 + (-5 - k)^2 = (\sqrt{65})^2\), which simplifies to \((3-h)^2 + (k + 5)^2 = 65\).

Step 5: Setting up the System of Equations

We now have a system of equations: 1) \(h^2 + (k + 6)^2 = 65\) 2) \((3-h)^2 + (k + 5)^2 = 65\).

Step 6: Simplifying and Solving the System of Equations

Expand and simplify both equations to solve for \(h\) and \(k\). Calculate the results using common algebraic methods.

Step 7: Finding the Result

Solving the two equations, we find two possible centers where both points can satisfy the circle equations. These coordinates put into respective circle equations will satisfy the mentioned radius requirement.

Key Concepts

System of Equations

A system of equations involves solving multiple equations at the same time to find the values that satisfy all of them. In this exercise, we have two equations from the circle equation

• \(h^2 + (k + 6)^2 = 65\)
• \((3-h)^2 + (k + 5)^2 = 65\)

To solve these simultaneously, you must find values for \(h\) and \(k\) that satisfy both equations.
This involves algebraic manipulation, like expanding, substituting, or using methods such as substitution or elimination.
The objective is to isolate the variables and solve for their values, leading us to the centers of the desired circles.

Radius of a Circle

The radius of a circle is the distance from the center of the circle to any point on the circle. In our problem, the radius is given as \(\sqrt{65}\).
This tells us that every point along the edge of our circles is \(\sqrt{65}\) units away from the center.
When given this value, it becomes a crucial component in forming our circle equations.
The equation of a circle is

• \((x - h)^2 + (y - k)^2 = r^2\)

where \((h, k)\) is the center of the circle and \(r\) is the radius.
This equation allows us to verify or find any point along the circumference when the radius and center are known.

Circle Centers

Finding the center of a circle involves determining the coordinate \((h, k)\) that makes the circle equation

• \((x - h)^2 + (y - k)^2 = r^2\)

true for given points on the circle’s edge.
In this task, we used two given points

• \((0, -6)\)
• \((3, -5)\)

to create our system of equations and determine the center coordinates.
Each point satisfies the circle equation when substituted for \((x, y)\), allowing you to create separate equations that both equal \(r^2\), the square of the radius.
By solving this system, you find potential circle centers.

Coordinate Geometry

Coordinate geometry, or analytic geometry, involves the use of coordinate points to solve geometry problems. In our exercise, we used it to find circle centers based on given circle features like radius and points.
For circle equations, coordinate geometry helps us comprehend and calculate relationships and measurements in a coordinate plane.
It allows for graphical representation of geometric entities through equations and coordinates.
By applying algebraic methods within this framework, one can precisely identify shapes, distances, and positional relationships on a plane.
This is especially advantageous for problems involving circles, as their geometric properties relate harmoniously with algebraic equations.