Problem 56
Question
Determine a value of \(k\) for which the graph of the given Cartesian equation is a point. \(x^{2}-6 x+y^{2}+2 y+k=1\)
Step-by-Step Solution
Verified Answer
The value of \(k\) is 10.
1Step 1: Identify the Form of a Circle Equation
To determine when the equation represents a point, we need to recognize it can be a circle's equation if it's of the form: \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center, and \(r\) is the radius. A point occurs when \(r^2 = 0\).
2Step 2: Complete the Square for x and y Terms
Rewrite the equation to complete the square: \(x^2 - 6x + y^2 + 2y + k = 1\).For \(x^2 - 6x\), use \((x - 3)^2 - 9\) (since \(-6/2=-3\), and \((-3)^2 = 9\)).For \(y^2 + 2y\), use \((y + 1)^2 - 1\) (since \(2/2=1\), and \(1^2 = 1\)).
3Step 3: Substitute Completed Squares into the Equation
The equation becomes:\((x-3)^2 - 9 + (y+1)^2 - 1 + k = 1\).Simplify to:\((x-3)^2 + (y+1)^2 + k - 10 = 1\).
4Step 4: Set the Equation to Represent a Point
For a circle reduced to a point, the equation must satisfy \((x-h)^2 + (y-k)^2 = 0\).Thus, set the constant term on the right side to zero:\(k - 10 = 0\).
5Step 5: Solve for k
Solve for \(k\) in the equation:\(k - 10 = 0\) implies \(k = 10\).
Key Concepts
Completing the SquareEquation of a CircleGraph of a Point
Completing the Square
Completing the square is a fundamental technique used in algebra to transform quadratic expressions into a form that's easier to analyze and work with. Let's dive into an example to understand how it works. When we look at an expression like \(x^2 - 6x\), our goal is to express it as a perfect square trinomial. This means we want to rewrite it as something like \((x - a)^2\).
To do this:
The same technique is applied to the \(y\)-terms in the equation. Take \(y^2 + 2y\), divide the \(y\)-term's coefficient by 2 to get 1, then square it to obtain 1. Add and subtract 1 to get \((y+1)^2 - 1\).
Completing the square is particularly useful in converting general equations into specific forms, like transforming a standard quadratic equation into the canonical form of a circle. This simplification helps make analyses and calculations more straightforward.
To do this:
- Take the coefficient of the \(x\)-term, which is \(-6\). Divide it by 2, resulting in \(-3\).
- Square this result to get 9.
The same technique is applied to the \(y\)-terms in the equation. Take \(y^2 + 2y\), divide the \(y\)-term's coefficient by 2 to get 1, then square it to obtain 1. Add and subtract 1 to get \((y+1)^2 - 1\).
Completing the square is particularly useful in converting general equations into specific forms, like transforming a standard quadratic equation into the canonical form of a circle. This simplification helps make analyses and calculations more straightforward.
Equation of a Circle
The equation of a circle provides a concise way to represent the set of all points that are equidistant from a given point, called the center. A circle is commonly expressed in the form \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) are the coordinates of the center, and \(r\) is the radius.
In the context of our problem, we started with a complex quadratic expression and used completing the square to transform it to something resembling the circle's equation. After completing the square, we obtained terms like \((x-3)^2\) and \((y+1)^2\). These give us hints about the center of the circle with \(h = 3\) and \(k = -1\).
This form directly indicates the location of the circle's center.
When the problem asks for the graph to represent a single point, we're looking at the case where the radius \(r\) is zero. This zero-radius means that every point on the circle is the same point, thus reducing it to a single point. Therefore, when solving these types of exercises, converting to this circle form can reveal critical information about the graph's nature.
In the context of our problem, we started with a complex quadratic expression and used completing the square to transform it to something resembling the circle's equation. After completing the square, we obtained terms like \((x-3)^2\) and \((y+1)^2\). These give us hints about the center of the circle with \(h = 3\) and \(k = -1\).
This form directly indicates the location of the circle's center.
When the problem asks for the graph to represent a single point, we're looking at the case where the radius \(r\) is zero. This zero-radius means that every point on the circle is the same point, thus reducing it to a single point. Therefore, when solving these types of exercises, converting to this circle form can reveal critical information about the graph's nature.
Graph of a Point
Visualizing a point on a graph is much simpler than handling complex shapes like circles or parabolas. A point, in its essence, is just a location in the coordinate system without any dimension – it has no width or area.
To represent a point using an equation, such as changing a circle equation to just a point, we rely on the circle's radius condition.
For a circle, the equation \((x-h)^2 + (y-k)^2 = r^2\) becomes a point at its center when \(r = 0\). Thus, the common term becomes \((x-h)^2 + (y-k)^2 = 0\), meaning:\
In the exercise, solving for \(k=10\) allowed us to set the remaining terms of the equation to zero, ensuring that the graph is just the point \((3, -1)\), where the circle collapses into a single dot on the Cartesian plane.
To represent a point using an equation, such as changing a circle equation to just a point, we rely on the circle's radius condition.
For a circle, the equation \((x-h)^2 + (y-k)^2 = r^2\) becomes a point at its center when \(r = 0\). Thus, the common term becomes \((x-h)^2 + (y-k)^2 = 0\), meaning:\
- \(x = h\)
- \(y = k\)
In the exercise, solving for \(k=10\) allowed us to set the remaining terms of the equation to zero, ensuring that the graph is just the point \((3, -1)\), where the circle collapses into a single dot on the Cartesian plane.
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