The x intercept of the circle `x^(2)+y^(2)+8x-9=0` is

Options

A. 8
B. 1
C. 7
D. 20

Answer

To find the x-intercepts of the circle given by the equation \( x^2 + y^2 + 8x - 9 = 0 \), we follow these steps:

### Step 1: Set \( y = 0 \)
Since we are looking for the x-intercepts, we need to find the points where the circle intersects the x-axis. On the x-axis, the value of \( y \) is always \( 0 \). Therefore, we substitute \( y = 0 \) into the equation of the circle.

### Step 2: Substitute \( y = 0 \) into the equation
Substituting \( y = 0 \) into the circle's equation:
\[ x^2 + 0^2 + 8x - 9 = 0 \]
This simplifies to:
\[ x^2 + 8x - 9 = 0 \]

### Step 3: Factor the quadratic equation
Now we need to factor the quadratic equation \( x^2 + 8x - 9 = 0 \). We look for two numbers that multiply to \(-9\) and add to \(8\). The numbers \(9\) and \(-1\) satisfy these conditions.

Thus, we can factor the equation as:
\[ (x + 9)(x - 1) = 0 \]

### Step 4: Solve for \( x \)
Setting each factor to zero gives us the x-values:
1. \( x + 9 = 0 \) → \( x = -9 \)
2. \( x - 1 = 0 \) → \( x = 1 \)

### Step 5: State the x-intercepts
The x-intercepts of the circle are the points where the circle intersects the x-axis. Therefore, the x-intercepts are:
\[ (-9, 0) \quad \text{and} \quad (1, 0) \]

### Summary
The x-intercepts of the circle \( x^2 + y^2 + 8x - 9 = 0 \) are \( x = -9 \) and \( x = 1 \).

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