Question

Graph each equation and find the point(s) of intersection, if any. The circle \((x+2)^{2}+(y-1)^{2}=4\) and the parabola \(y^{2}-2 y-x-5=0\)

Short answer

The points of intersection are the values of \((x, y)\) after substituting solutions in the circle and solving the system. Solve \( (x+2)^{2}+(y-1)^{2}=4 \) with \) y^{2}-2 y-x-5=0 $.

Step-by-step solution

Understand the Equations

The first equation is a circle with center \text{(-2, 1)} and radius 2: \((x+2)^{2}+(y-1)^{2}=4\). The second equation is a parabola with its standard form being rewritten as \(y^2 - 2y = x + 5\).

Rewrite the Parabola Equation

Rewrite the equation for the parabola to isolate \(x\): \[ x = y^2 - 2y - 5 \] We'll graph this parabola in this form.

Solve the Circle Equation in Terms of x

Rearrange the circle equation: \[ (x+2)^2+(y-1)^2 = 4 \] Expand and simplify: \[ x^2 + 4x + 4 + y^2 - 2y + 1 = 4 \]Combining like terms and simplifying: \[ x^2 + 4x + y^2 - 2y + 1 = 0 \]

Set the Equations Equal

Substitute the expression for \(x\) from the parabola into the equation for the circle: \[ (y^2 - 2y - 5 + 2)^2 + (y - 1)^2 = 4 \]Simplify it step by step to solve for \(y\).

Solve for y

Simplify further: \[ ((y^2 - 2y - 3)^2 + (y - 1)^2 = 4) \] This results in a polynomial equation. Solve for \(y\).

Find Corresponding x Values

Using the values of \(y\) obtained from the polynomial equation, substitute back to find \(x\). This will give the points of intersection.

Key concepts

graphing circles

The first concept involves graphing a circle. A circle's equation is typically written in the form \((x-h)^2 + (y-k)^2 = r^2\) where \( (h,k) \) is the center of the circle and \( r \) is its radius. For instance, the equation \((x+2)^2 + (y-1)^2 = 4\)\, translates to a circle with center \((-2, 1)\) and radius 2.

To graph this circle, follow these steps:

• Locate the center point on the coordinate plane at \((-2, 1)\).
• From the center, measure a distance of 2 units (the radius) in all directions.
• Mark these points and draw the circle.

Understanding this foundation is key for visualizing the intersection with other curves.

solving polynomial equations

A polynomial equation consists of variables and coefficients combined using addition, subtraction, and multiplication. In this exercise, the polynomial equation results from setting the circle's equation equal to the parabola's equation.

For example, simplifying the circle's equation \( (y^2 - 2y - 3)^2 + (y-1)^2 = 4\)leads to a polynomial equation in terms of \( y \). Here's how to solve:

• Expand and simplify the polynomial expression.
• Combine like terms.
• Factorize or use the quadratic formula to obtain the solutions for \( y \).

By solving for \( y \), we can find the corresponding \( x \) values, giving us the intersection points.

intersection of curves

Finding the points where two curves intersect involves solving simultaneous equations. In this case, the circle and parabola equations are: \((x+2)^2 + (y-1)^2 = 4\)\ and \(y^2 - 2y = x + 5\). Substitute the expression \(x = y^2-2y-5\)\ into the circle's equation.
Then, simplify:
\[((y^2-2y-3)^2 + (y-1)^2 = 4)\]

• Unravel the left side to make it a polynomial in \(y\).
• Solve the polynomial for \(y\).
• Substitute back to get the corresponding \( x \) values.

The coordinate pairs \((x,y)\)\ provide the points where the curves intersect.

graphing parabolas

A parabola can be represented in the form \(y = ax^2 + bx + c\)\ or similar. In this exercise, the parabola equation: \(y^2 - 2y = x + 5\)\ is manipulated to \(x = y^2 - 2y - 5\).

To graph this parabola:

• Identify the vertex, which is at the minimum or maximum point of the parabola.
• Plot key points on either side of the vertex to ascertain the curve's shape.
• Remember that parabolas open upwards/downwards depending on the coefficient sign of the squared term.

By following these steps, you can accurately graph a parabola.

coordinates

Coordinates help us pinpoint exact locations on a graph. Each point is represented as \((x, y)\).

For instance, when we find the intersection points, we calculate \(y\) values first, then, using those \(y\) values, find the respective \(x\) values.
These coordinate pairs \((x, y)\)\ show where the circle and parabola meet. It's essential to accurately calculate and plot these points on a coordinate plane for clear visualization.

• Identify key points (intercepts, vertices, etc.)
• Ensure scales are consistent on both axes.

Understanding how to work with coordinates is crucial for graphing and solving conic sections.