Show that `x + y + 1 = 0` touches the circle
`x^(2) + y^(2) -3x + 7y +14 = 0` and find its
point of contact.

Options

A. a) (1,0)
B. b) (2,-3)
C. c) (5,3)
D. d) (-1,0)

Answer

To show that the line \( x + y + 1 = 0 \) touches the circle given by the equation \( x^2 + y^2 - 3x + 7y + 14 = 0 \) and to find the point of contact, we can follow these steps:

### Step 1: Rewrite the line equation
The line equation can be rewritten to express \( y \) in terms of \( x \):
\[ y = -x - 1 \]

**Hint:** Isolate \( y \) in the line equation to substitute it into the circle's equation.

### Step 2: Substitute \( y \) into the circle's equation
Now, substitute \( y = -x - 1 \) into the circle's equation:
\[ x^2 + (-x - 1)^2 - 3x + 7(-x - 1) + 14 = 0 \]

### Step 3: Simplify the equation
Expanding \( (-x - 1)^2 \):
\[ (-x - 1)^2 = x^2 + 2x + 1 \]
Now substituting this back into the equation:
\[ x^2 + (x^2 + 2x + 1) - 3x - 7x - 7 + 14 = 0 \]
Combining like terms:
\[ 2x^2 + 2x + 1 - 10x + 7 = 0 \]
This simplifies to:
\[ 2x^2 - 8x + 8 = 0 \]
Dividing the entire equation by 2:
\[ x^2 - 4x + 4 = 0 \]

### Step 4: Factor the quadratic equation
The equation can be factored as:
\[ (x - 2)^2 = 0 \]
This gives us:
\[ x = 2 \]

**Hint:** Look for perfect squares when factoring to find the roots.

### Step 5: Find the corresponding \( y \) value
Now substitute \( x = 2 \) back into the equation for \( y \):
\[ y = -2 - 1 = -3 \]

### Step 6: State the point of contact
Thus, the point of contact between the line and the circle is:
\[ (2, -3) \]

### Conclusion
We have shown that the line \( x + y + 1 = 0 \) touches the circle \( x^2 + y^2 - 3x + 7y + 14 = 0 \) at the point \( (2, -3) \).