Problem 70 - Algebra 2

Question

a. A circle contains \((0,0),(6,8),\) and \((7,7)\) . Find its equation by solving a system of three equations. b. Several parabolas contain the three points of part (a), but only one is described by a quadratic function. Find that function.

Step-by-Step Solution

Answer

The equation for the circle is \(x^2 + y^2 - 1.2x - 0.8y = 0\) and for the parabola is \(y=-0.2x^2+2.6x\).

Step 1: Equation of a Circle

The general equation of a circle is given by \(x^2 + y^2 + 2gx + 2fy + c = 0\). Assume (0,0), (6,8), (7,7) lie on the circle, substitute these points into the equation to get three equations. 1) For (0,0), the equation reduces to \(c=0\). 2) For (6,8), plug these values into the equation to get \(36 + 64 + 12g + 16f +0 = 0\) which simplifies to \(6g + 8f = -10\) or \(3g+4f=-5\). 3) For (7,7), we get \(49 + 49 + 14g + 14f + 0 = 0\) which simplifies to \(7g+7f=-7\) or \(g+f=-1\). We have three equations: \(c=0\), \(3g+4f=-5\), and \(g+f=-1\).

Step 2: Solving the system of equations

We can solve the system of equations \(3g+4f=-5\) and \(g+f=-1\) simultaneously. By multiplying equation 2 by 3 and subtracting equation 3 from it, we get \(5f=-2\) or \(f=-0.4\). Plugging \(f\) into equation 3, we find \(g=-0.6\). So, the equation for the circle is \(x^2 + y^2 - 1.2x - 0.8y = 0\).

Step 3: Equation of a Parabola

The general equation of a parabola is \(y=ax^2+bx+c\). Substitute each point into the equation to get a system of equations: 1) For (0,0), we find \(c=0\). 2) For (6,8), we find \(8=36a+6b+0\). 3) For (7,7), we get \(7=49a+7b\). This gives: \(c=0\), \(36a+6b=8\), and \(49a+7b=7\).

Step 4: Solving the system of equations

Let's solve the system of equations \(36a+6b=8\) and \(49a+7b=7\) simultaneously. Subtract the second equation from the first to get \(-13a=b\). Plugging \(b=-13a\) into to equation 2, we get \(36a-78a=8\) which gives \(a=-0.2\). Therefore, \(b=2.6\). So, the equation for the parabola is \(y=-0.2x^2+2.6x\).

Key Concepts

Equation of a Circle

The equation of a circle is a fundamental part of Analytic Geometry. It’s represented by the general formula:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
- Here, \(g\) and \(f\) are related to the circle's center coordinates, and \(c\) is a constant.
To determine the specific equation of a circle when given three points, you substitute these points into the general equation to form a system of linear equations. This process will help you find the values of \(g\), \(f\), and \(c\).

• For the origin \( (0,0) \), substitution gives \( c = 0 \).
• Substituting points \( (6,8) \) or \( (7,7) \), gives two more equations that are combined to solve for \(g\) and \(f\).

This approach allows us to derive the particular equation of the circle using provided coordinate points. Becoming comfortable with this derivation is a stepping stone into more complex concepts within Analytic Geometry.

Quadratic Functions

Quadratic functions are polynomials of degree two and take the form:
\[ y = ax^2 + bx + c \]
In this equation, \(a\), \(b\), and \(c\) are constants, where \(a eq 0\), and they determine the shape and position of the parabola on a graph.

• The coefficient \(a\) affects the parabola's direction (upward if \(a > 0\) or downward if \(a < 0\)).
• The vertex form can additionally help reveal the peak or trough point of the curve.

When solving for the specific quadratic function that passes through three points, you align each point with the equation to generate a system of equations. Solving these simultaneously allows you to pinpoint the values of \(a\), \(b\), and \(c\).
This process is crucial for finding the unique quadratic function amidst several possibilities that meet certain required conditions.

System of Equations

A system of equations comprises multiple equations that are considered simultaneously. The objective is to find common solutions to all equations in the system. Systems can be linear or non-linear, involving variables and constants. In the context of this exercise:

• A system was formed using the coordinates of points to determine the equation of a circle, where each point provided a linear equation derived from substituting into the circle's general formula.
• Another system was derived for the parabola, using the quadratic function equation to give a quadratic curve passing through the same points.

Solving these systems involves methods such as substitution, elimination, or matrix operations. For linear systems, simplification often involves manipulating equations to isolate variables so they can be solved sequentially.
Understanding and solving systems of equations is a valuable skill in both math and real-life applications, helping to determine unknowns that define specific conditions or standards.