Problem 61 - Contemporary Precalculus

Question

Sketch the graph of the equation. Label the \(x\) - and y-intercepts. $$(x+1)^{2}+(y-3)^{2}=9$$

Step-by-Step Solution

Answer

Answer: The center of the circle is at \((-1, 3)\), the radius is \(3\), the x-intercept is \((-1, 0)\), and the y-intercepts are \((0, 3+\sqrt{8})\) and \((0, 3-\sqrt{8})\).

Step 1: Identify the circle's properties

Comparing the equation \((x+1)^2+(y-3)^2=9\) with the standard form \((x-a)^2 + (y-b)^2 = r^2\), we can see that \(a = -1\), \(b = 3\), and \(r^2 = 9\). Therefore, the center of the circle is at \((-1, 3)\), and the radius is \(r = 3\).

Step 2: Find the x-intercepts

To find the x-intercepts, set \(y = 0\) in the equation, and solve for \(x\): $$(x+1)^2 + (0-3)^2 = 9$$ $$(x+1)^2 + 9 = 9$$ $$(x+1)^2 = 0$$ Therefore, \(x = -1\) is the only x-intercept, which is the point \((-1,0)\).

Step 3: Find the y-intercepts

To find the y-intercepts, set \(x = 0\) in the equation, and solve for \(y\): $$(0+1)^2 + (y-3)^2 = 9$$ $$1 + (y-3)^2 = 9$$ $$(y-3)^2 = 8$$ Taking the square root of both sides, we get \(y-3 = \pm \sqrt{8}\). Therefore, there are two \(y\)-intercepts: \(y = 3+\sqrt{8}\) and \(y = 3-\sqrt{8}\). These are the points \((0,3+\sqrt{8})\) and \((0,3-\sqrt{8})\).

Step 4: Sketch the graph

To sketch the graph, plot the center of the circle at \((-1, 3)\), and draw a circle with a radius of \(3\). Label the x-intercept at \((-1, 0)\) and the y-intercepts at \((0, 3+\sqrt{8})\) and \((0, 3-\sqrt{8})\). The graph represents a circle with these intercepts.

Key Concepts

Precalculus

In precalculus, we delve into various mathematical concepts and techniques used to prepare for calculus, including understanding the fundamental aspects of geometric shapes such as circles. The equation \[(x+1)^2 + (y-3)^2 = 9\]is a specific case of a circle equation in precalculus. Understanding this helps set the foundation for more advanced topics in calculus and analytical geometry.

In this context, the equation of the circle is derived from the general form \[(x-a)^2 + (y-b)^2 = r^2\], where \(a\) and \(b\) are coordinates for the center and \(r\) represents the radius. This representation helps us visualize and solve problems related to circles in a systematic way.

• The equation is used to determine the center, which is clearly identified as \((-1, 3)\).
• The radius of 3 is derived from the square root of 9, as it's the squared term \(r^2\) that equals 9.

These insights allow students to logically deduce properties directly from the circle's equation, making future geometric and calculus problems more approachable.

Coordinate Geometry

Coordinate geometry, often known as analytic geometry, is the study of geometry using a coordinate system. This branch of geometry allows us to apply algebraic principles to geometric problems, merging these concepts seamlessly. With circles, coordinate geometry enables the conversion of geometric shapes into algebraic equations that describe their properties.

• By analyzing the equation of the circle, \[(x+1)^2 + (y-3)^2 = 9,\]we identify the parameters of the circle, such as the center \((-1, 3)\) and the radius 3.
• The problem-solving approach using intercepts highlights the application of algebra in geometry. For x-intercepts, setting \(y = 0\) allows us to find \(x = -1\). For y-intercepts, setting \(x = 0\) gives us solutions in terms of \(y = 3 \pm \sqrt{8}\).

These techniques illustrate how coordinate geometry facilitates finding specific points where the circle intersects the axes, which is essential for graphing the shape accurately.

Graphing Circles

Graphing circles is a fundamental skill in coordinate geometry and precalculus that requires identifying key components of the circle from its equation and plotting them accurately. The standard form of a circle's equation, \[(x-a)^2 + (y-b)^2 = r^2,\]gives us the necessary elements to draw the circle accurately on the Cartesian plane.
The exercise demands we sketch the circle by placing its center at \((-1, 3)\) based on the simplified equation \[(x+1)^2 + (y-3)^2 = 9.\]The radius specifies how far from the center the circle extends, generating a round shape.

• Graphing involves marking crucial points like the intercepts \((-1,0)\), and \((0,3\pm\sqrt{8})\) to form an exact representation of the circle.
• These calculations align with the visualization, ensuring accuracy when transitioning from the equation to a visual representation.

By correlating numeric outcomes with graph points, graphing circles confirms knowledge and offers a graphical perspective of abstract numerical relationships. This preparation equips students with the skills required for more complex graphing tasks that will appear later in their studies.