Question

X^2+y^2-8x+6y+21=0 решите и нарисуйте график Помогите пожалуйста

Answer 1

Quadratic Equation and Graph Plotting

To solve the equation x^2 + y^2 - 8x + 6y + 21 = 0 and plot its graph, we can follow these steps:

1. Complete the square for both the x and y terms to convert the equation into the standard form of a circle equation. 2. Identify the center and radius of the circle. 3. Plot the circle on a graph.

Let's go through each step in detail.

Step 1: Completing the Square

To complete the square for the x terms, we need to add and subtract the square of half the coefficient of x. Similarly, for the y terms, we add and subtract the square of half the coefficient of y.

The equation x^2 + y^2 - 8x + 6y + 21 = 0 can be rewritten as:

(x^2 - 8x) + (y^2 + 6y) + 21 = 0

To complete the square for the x terms, we add and subtract (8/2)^2 = 16:

(x^2 - 8x + 16) + (y^2 + 6y) + 21 - 16 = 0

To complete the square for the y terms, we add and subtract (6/2)^2 = 9:

(x^2 - 8x + 16) + (y^2 + 6y + 9) + 21 - 16 - 9 = 0

Simplifying further:

(x - 4)^2 + (y + 3)^2 + 16 = 0

Step 2: Identifying the Center and Radius

Comparing the equation with the standard form of a circle equation:

(x - h)^2 + (y - k)^2 = r^2

We can identify the center (h, k) and radius r.

In our equation, (x - 4)^2 + (y + 3)^2 + 16 = 0, the center is (4, -3) and the radius is √16 = 4.

Step 3: Plotting the Circle

Now that we have the center and radius, we can plot the circle on a graph.

The center of the circle is (4, -3), and the radius is 4. We can plot the center as a point on the graph and draw a circle with a radius of 4 around it.

Please note that the equation (x - 4)^2 + (y + 3)^2 + 16 = 0 represents a degenerate circle, which means it does not intersect the x-y plane. Therefore, the graph will be a single point at the center (4, -3).

I hope this helps! Let me know if you have any further questions.