Problem 17
Question
In Exercises 15 through 20 , determine whether the graph is a circle, a point- circle, or the empty set. $$ x^{2}+y^{2}-10 x+6 y+36=0 $$
Step-by-Step Solution
Verified Answer
The graph is the empty set.
1Step 1 - Identify general form
Recognize that the given equation is in the general form: \[ x^2 + y^2 + Dx + Ey + F = 0 \]For this problem: \[ x^2 + y^2 - 10x + 6y + 36 = 0 \]Thus, \(D = -10\), \(E = 6\), and \(F = 36\).
2Step 2 - Complete the square for x
To complete the square for the \(x\) terms, take half of the coefficient of \(x\) (which is -10), square it, and add and subtract this value inside the equation. \(\left( \frac{-10}{2} \right)^2 = (-5)^2 = 25\)Adding and subtracting 25 gives: \[ x^2 - 10x + 25 - 25 \]Rewrite this as: \[ (x - 5)^2 - 25 \]
3Step 3 - Complete the square for y
To complete the square for the \(y\) terms, take half of the coefficient of \(y\) (which is 6), square it, and add and subtract this value inside the equation. \(\left( \frac{6}{2} \right)^2 = 3^2 = 9\)Adding and subtracting 9 gives: \[ y^2 + 6y + 9 - 9 \]Rewrite this as: \[ (y + 3)^2 - 9 \]
4Step 4 - Rewrite the equation
Substitute the completed squares for \(x\) and \(y\) back into the equation: \[ (x - 5)^2 - 25 + (y + 3)^2 - 9 + 36 = 0 \]
5Step 5 - Simplify the equation
Combine constants on the right side of the equation: \[ (x - 5)^2 + (y + 3)^2 - 25 - 9 + 36 = 0 \]\[ (x - 5)^2 + (y + 3)^2 + 2 = 0 \]Subtract 2 from both sides: \[ (x - 5)^2 + (y + 3)^2 = -2 \]
6Step 6 - Determine the nature of the graph
Since the right side of the equation \[ (x - 5)^2 + (y + 3)^2 = -2 \] is negative, it indicates that there is no real solution. Therefore, the graph represents the empty set.
Key Concepts
Circle EquationsGraph AnalysisCoordinate Geometry
Circle Equations
In coordinate geometry, a circle's equation in the standard form is written as \[(x - h)^2 + (y - k)^2 = r^2\], where *(h, k)* is the center of the circle and \(r\) is the radius. This equation is obtained by completing the square for the terms involving x and y.For example, consider an equation like \[x^2 + y^2 - 10x + 6y + 36 = 0\]. To transform this into standard form, first complete the square for the x terms and y terms separately. The completed squares reveal the center and radius. However, if you end up with a negative value on the right side of the equation, like -2 in this problem, it means the graph represents an empty set, because it's impossible to have a negative radius squared.
Graph Analysis
Graph analysis involves understanding the nature of the graph from its equation. For circle equations, a crucial aspect is determining whether the equation represents a real circle, a point-circle, or the empty set.
- A real circle occurs if the equation matches the form \[(x - h)^2 + (y - k)^2 = r^2\] with a positive r.
- A point-circle represents a degenerate case where the radius r is zero, reducing the equation to \[(x - h)^2 + (y - k)^2 = 0\]. This signifies a single point (h, k).
- The empty set happens if the right-hand side is negative, as in \[(x - 5)^2 + (y + 3)^2 = -2\]. No real values of x and y will satisfy this equation, so no graph exists.
Coordinate Geometry
Coordinate geometry is the study of geometry using a coordinate system. Here, we use the Cartesian coordinate system where each point on the plane is identified by an ordered pair (x, y).
- To derive the circle's equation from its general form \[Ax^2 + Ay^2 + Dx + Ey + F = 0\], complete the square for both x and y terms.
- The process of completing the square helps convert an equation into a recognizable geometric form.
- The coordinates of the center and the radius of a circle can be identified from the completed square form: \[(x - h)^2 + (y - k)^2 = r^2\].
Other exercises in this chapter
Problem 16
$$ \text { Find an equation of the line through the points }(3,-5) \text { and }(1,-2), \text { and put the equation in the slope-intercept form. } $$
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In Exercises 11 through 34, the function is the set of all ordered pairs \((x, y)\) satisfying the given equation. Find the domain and range of the function, an
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In Exercises 15 through 26 , find the solution set of the given inequality, and illustrate the solution on the real number line. $$ |3 x-4| \leq 2 $$
View solution Problem 17
Find the coordinates of the three points that divide the line segment from \(A(-5,3)\) to \(B(6,8)\) into four equal parts.
View solution