Problem 78 - Thomas Calculus

Question

Find an equation for the circle through the points \((2,3),(3,2),\) and \((-4,3) .\)

Step-by-Step Solution

Answer

The equation of the circle is \((x + 4)^2 + (y + 4.5)^2 = 92.25\).

Step 1: Identify the general equation of a circle.

The general equation of a circle with center \((h, k)\) and radius \(r\) is given by: \[(x - h)^2 + (y - k)^2 = r^2\] Our task is to find \(h\), \(k\), and \(r\) so that the circle passes through the given points.

Step 2: Set up the system of equations using the given points.

Substitute each point into the general circle equation to create a system of equations:For \((2,3)\): \[ (2 - h)^2 + (3 - k)^2 = r^2 \]For \((3,2)\): \[ (3 - h)^2 + (2 - k)^2 = r^2 \]For \((-4,3)\): \[ (-4 - h)^2 + (3 - k)^2 = r^2 \]

Step 3: Expand the equations and solve for \(h\) and \(k\).

Begin by expanding each equation:1. \((2-h)^2 + (3-k)^2 = r^2\) becomes \((h-2)^2 + (k-3)^2 = r^2\).2. \((3-h)^2 + (2-k)^2 = r^2\) becomes \((h-3)^2 + (k-2)^2 = r^2\).3. \((-4-h)^2 + (3-k)^2 = r^2\) becomes \((h+4)^2 + (k-3)^2 = r^2\).

Step 4: Use elimination or substitution to find \(h\) and \(k\).

Subtract the first equation from the second, and the first equation from the third to eliminate \(r^2\):- Subtracting 1st from 2nd: \[(h-3)^2 - (h-2)^2 + (k-2)^2 - (k-3)^2 = 0\] - Subtracting 1st from 3rd: \[(h+4)^2 - (h-2)^2 + (k-3)^2 - (k-3)^2 = 0\]Simplifying gives:- \(-2h + 2k + 1 = 0\) (1)- \(12h = -48\) (2)From (2), solve for \(h\): \(h = -4\). Substitute \(h\) into equation (1):\[-2(-4) + 2k + 1 = 0\]\[8 + 2k + 1 = 0\]\[2k = -9\]\[k = -4.5\]

Step 5: Find the radius \(r\) using \((h, k)\).

Substitute \(h = -4\) and \(k = -4.5\) back into the equation of any point, for example \((2,3)\):\[(2 + 4)^2 + (3 + 4.5)^2 = r^2\]\[6^2 + 7.5^2 = r^2\]\[36 + 56.25 = r^2\]\[r^2 = 92.25\]

Step 6: Write the final equation of the circle.

Using the values of \(h\), \(k\), and \(r\), write the final equation:\[(x + 4)^2 + (y + 4.5)^2 = 92.25\]

Key Concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is a branch of geometry where algebra is used to describe and analyze geometric figures. It relies on the use of a coordinate system, typically defined by the x-axis and y-axis in a 2-dimensional plane. This allows for a precise mathematical description of shapes.

By using coordinates, each point on the geometric figure can be expressed as an ordered pair \(x, y\), enabling the calculation of distances and areas between and of shapes more efficiently. In the context of circles, the coordinates aid in pinpointing specific points on the circle, helping define its properties and equation.

In this exercise, we are dealing with a circle defined by its relationship to three specific points. Understanding coordinate geometry allows us to substitute these points directly into the circle's equation to discover its other properties, such as radius and center.

System of Equations

A system of equations is a collection of two or more equations with the same set of variables. In finding the equation for a circle, systems of equations are particularly useful to handle multiple conditions that must be simultaneously satisfied. This method allows us to solve for the unknowns by substituting values and using algebraic manipulation.

For example, when given the three points through which a circle passes, we substitute each set of coordinates into the circle's general equation. This results in a system comprised of equations all equaling a common term, \(r^2\), the square of the circle's radius.

The system is instrumental in equating different expressions of \(r^2\) to isolate and solve for the circle's center coordinates \(h\) and \(k\). By systematically solving and substituting through these equations, the circle's unique properties are uncovered.

Substitution Method

The substitution method is a popular technique for solving systems of equations. It involves solving one of the equations for one variable and then substituting that expression into the other equation(s). This helps in reducing complex systems to simpler, more manageable forms by eliminating variables one by one.

In this exercise, we solve the first equation for \(r^2\) in terms of \(h\) and \(k\) and then apply elimination to get simpler equations focusing on \(h\) and \(k\). By substituting the value of \(h\) found from equation (2) into equation (1), we streamline the computation for \(k\).

Thus, substitution plays a crucial role in breaking down the complex problem of finding a circle equation into simpler steps wherein each variable is handled with focused attention. This results in accurate solutions that lead to correctly determining the circle's unique characteristics.

Circle Properties

A circle is a critical geometric shape with properties defined by its radius \(r\), center \(h, k\), and its equation. The fundamental equation \((x - h)^2 + (y - k)^2 = r^2\) expresses all these properties concisely.

When a circle passes through specified points, its radius and center are directly influenced by these coordinates. By substituting the given points into the circle's equation, and solving the resulting system, we determine the center \((-4, -4.5)\) and the radius \(\sqrt{92.25}\).

Understanding the properties of circles not only includes knowledge of how to express them numerically, but also a comprehension of their geometric implications, such as symmetry and uniform distance from the center to every point on its circumference. This knowledge allows us to utilize the circle's equation to express other properties or to sketch its accurate representation geometrically.