Tangent to the parabola `y=x^(2)+6` at (1, 7) touches the circle `x^(2)+y^(2) +16x +12y+c=0` at the point

Options

A. (-6, -9)
B. (-13, -9)
C. (-6, -7)
D. (13, 7)

Answer

To solve the problem step by step, we will find the equation of the tangent to the parabola at the given point, substitute it into the equation of the circle, and then determine the value of \( c \) such that the tangent touches the circle at a point.

### Step 1: Find the equation of the tangent to the parabola

The equation of the parabola is given as:
\[ y = x^2 + 6 \]
To find the slope of the tangent at the point \( (1, 7) \), we first differentiate the equation of the parabola:
\[ \frac{dy}{dx} = 2x \]
At \( x = 1 \):
\[ \frac{dy}{dx} = 2(1) = 2 \]
The slope of the tangent line at the point \( (1, 7) \) is \( 2 \).

Using the point-slope form of the line, the equation of the tangent line can be written as:
\[ y - y_1 = m(x - x_1) \]
Substituting \( m = 2 \), \( (x_1, y_1) = (1, 7) \):
\[ y - 7 = 2(x - 1) \]
Simplifying this:
\[ y - 7 = 2x - 2 \implies y = 2x + 5 \]
This is our equation of the tangent line.

### Step 2: Substitute the tangent line equation into the circle's equation

The equation of the circle is given as:
\[ x^2 + y^2 + 16x + 12y + c = 0 \]
Substituting \( y = 2x + 5 \) into the circle's equation:
\[ x^2 + (2x + 5)^2 + 16x + 12(2x + 5) + c = 0 \]
Expanding \( (2x + 5)^2 \):
\[ x^2 + (4x^2 + 20x + 25) + 16x + (24x + 60) + c = 0 \]
Combining like terms:
\[ 5x^2 + 60x + 85 + c = 0 \]

### Step 3: Set the discriminant to zero

Since the tangent touches the circle, the discriminant of the quadratic equation must be zero:
\[ b^2 - 4ac = 0 \]
Here, \( a = 5 \), \( b = 60 \), and \( c = 85 + c \):
\[ 60^2 - 4 \cdot 5 \cdot (85 + c) = 0 \]
Calculating \( 60^2 \):
\[ 3600 - 20(85 + c) = 0 \]
Expanding:
\[ 3600 - 1700 - 20c = 0 \implies 1900 = 20c \implies c = \frac{1900}{20} = 95 \]

### Step 4: Find the point of tangency

Now, substituting \( c = 95 \) back into the quadratic equation:
\[ 5x^2 + 60x + 180 = 0 \]
We can find the roots using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Since the discriminant is zero, we have:
\[ x = \frac{-60}{2 \cdot 5} = -6 \]
Now substituting \( x = -6 \) back into the tangent line equation to find \( y \):
\[ y = 2(-6) + 5 = -12 + 5 = -7 \]

### Final Answer

The point of tangency is:
\[ (-6, -7) \]