Problem 55 - Precalculus

Question

Determine whether each statement is true or false. The circle given by the equation \(x^{2}+y^{2}+a x+b y+c=0\) passes through the points \((-2,4),(1,1),\) and \((-2,-2) .\) Find \(a, b,\) and \(c\)

Step-by-Step Solution

Answer

The coefficients are: \(a = 4\), \(b = -2\), \(c = -4\). The statements are true.

Step 1: General Equation of a Circle

The given equation is for a circle: \(x^2 + y^2 + ax + by + c = 0\). Every point \((x, y)\) that lies on this circle will satisfy this equation, meaning if we substitute \(x\) and \(y\) from the point into the equation, it will hold true.

Step 2: Substitute Points into the Equation

Substitute each of the given points into the equation:1. For the point \((-2, 4)\): \((-2)^2 + 4^2 + a(-2) + b(4) + c = 0\)2. For the point \((1, 1)\): \(1^2 + 1^2 + a(1) + b(1) + c = 0\)3. For the point \((-2, -2)\): \((-2)^2 + (-2)^2 + a(-2) + b(-2) + c = 0\).

Step 3: Solve the System of Equations

From Step 2, we obtain the following system of equations:1. \(4 + 16 - 2a + 4b + c = 0 \rightarrow -2a + 4b + c = -20\)2. \(1 + 1 + a + b + c = 0 \rightarrow a + b + c = -2\)3. \(4 + 4 - 2a - 2b + c = 0 \rightarrow -2a - 2b + c = -8\)We now solve this system to find \(a, b,\) and \(c\).

Step 4: Use Elimination or Substitution

Using elimination:1. Subtract equation 2 from equation 1: - \(-2a + 4b + c = -20\) - \( a + b + c = -2\) - This eliminates \(c,\) and results in: \(-3a + 3b = -18\) or \(-a + b = -6\)2. Subtract equation 2 from equation 3: - \(-2a - 2b + c = -8\) - \( a + b + c = -2\) - This eliminates \(c,\) and results in: \(-3a - 3b = -6\) or \(-a - b = -2\).

Step 5: Solve for a and b

We now solve the simplified system:1. \(-a + b = -6\)2. \(-a - b = -2\)By adding these equations, we eliminate \(b\): - \((-a + b) + (-a - b) = -6 + (-2)\) - \(-2a = -8\) - Solving gives: \(a = 4\).Now substitute \(a = 4\) into one of the simplified equations, say \(-a + b = -6\): - \(-4 + b = -6\) - Solving gives: \(b = -2\).

Step 6: Solve for c using a and b

Substitute \(a = 4\) and \(b = -2\) into one of the original equations, e.g., \(a + b + c = -2\): \(4 - 2 + c = -2\)Solve for \(c\): - \(2 + c = -2\) - \(c = -4\).Thus, we have found \(a = 4\), \(b = -2\), and \(c = -4\).

Key Concepts

system of equations

A system of equations is a collection of two or more equations with the same set of variables. In this problem, we used a system of three equations derived from substituting the given points into the circle's equation. By doing this, each equation becomes an expression of the relationship between the unknowns: \(a\), \(b\), and \(c\).
The goal is to solve for these unknowns using all three equations together.

• Equation 1: \(-2a + 4b + c = -20\)
• Equation 2: \(a + b + c = -2\)
• Equation 3: \(-2a - 2b + c = -8\)

Solving a system of equations means finding the values of the unknowns that satisfy all equations at the same time. The most common methods used are substitution and elimination.

point substitution

Point substitution refers to replacing the variables \(x\) and \(y\) in the circle's equation with specific values from the given points, ensuring these values satisfy the equation.
For example, the point \((-2, 4)\) is substituted into the equation \(x^2 + y^2 + ax + by + c = 0\) to form one equation. Each point results in a separate equation:

• For \((-2, 4)\): \((-2)^2 + 4^2 - 2a + 4b + c = 0\)
• For \((1, 1)\): \(1^2 + 1^2 + a + b + c = 0\)
• For \((-2, -2)\): \((-2)^2 + (-2)^2 - 2a - 2b + c = 0\)

By substituting the coordinates directly into the equation, we form a system of linear equations in terms of \(a\), \(b\), and \(c\). This method confirms that each point lies on the circle if the values of \(a\), \(b\), and \(c\) hold true for these substitutions.

elimination method

The elimination method is a strategy for solving systems of equations. It involves adding or subtracting equations to eliminate one variable, making it easier to solve the remaining equations.
In this exercise, the elimination method was used to find \(a\) and \(b\) by eliminating \(c\) from pairs of equations. Here's how it works:

• First, subtract equation 2 from equation 1 to eliminate \(c\). The result is \(-3a + 3b = -18\), which simplifies to \(-a + b = -6\).
• Then, subtract equation 2 from equation 3 to eliminate \(c\) again. This results in \(-3a - 3b = -6\), simplifying to \(-a - b = -2\).
• By adding these two new equations, \(b\) gets eliminated: \(-a + b + -a - b = -8\) simplifies to \(-2a = -8\). Solving gives \(a = 4\).

The value of \(a\) is then used to substitute back into one of the simplified equations, eventually solving for \(b\) and \(c\). This systematic approach is particularly useful in linear equations.

circle geometry

Circle geometry helps us understand the properties and equations of circles. A circle is a set of points that are equidistant from a central point, known as the center. The standard form of a circle's equation in the context of this problem looks slightly different with coefficients, which is given as: \(x^2 + y^2 + ax + by + c = 0\).
Here are some essential points about this type of circle equation:

• \(x^2\) and \(y^2\) denote the squared coordinates, representing the distance from the origin.
• The terms \(ax\) and \(by\) represent the linear adjustments needed to center the circle at a point other than the origin.
• The constant \(c\) shifts the entire circle in space, affecting its position on the graph.

Through solving this exercise,—finding \(a\), \(b\), and \(c\)—one deduces how these coefficients adjust the circle's properties in the coordinate plane, confirming the points it passes through.