Equation of tangent to the circle `x^(2)+y^(2)=50` at the point where the line `x+7=0` meets the circle

Options

A. `7x+y=50`
B. `x+7y=50`
C. `x pm 7y=50`
D. `7x pm y=50`

Answer

To find the equation of the tangent to the circle \( x^2 + y^2 = 50 \) at the point where the line \( x + 7 = 0 \) meets the circle, we can follow these steps:

### Step 1: Find the intersection point of the line and the circle.
The line \( x + 7 = 0 \) can be rewritten as:
\[ x = -7 \]

Now, substitute \( x = -7 \) into the equation of the circle:
\[ (-7)^2 + y^2 = 50 \]
This simplifies to:
\[ 49 + y^2 = 50 \]
Subtracting 49 from both sides gives:
\[ y^2 = 1 \]
Taking the square root of both sides, we find:
\[ y = \pm 1 \]

Thus, the points of intersection are:
\[ (-7, 1) \quad \text{and} \quad (-7, -1) \]

### Step 2: Use the point of tangency to find the tangent line.
We will use the point \((-7, 1)\) to find the equation of the tangent line. The general formula for the tangent to the circle \( x^2 + y^2 = r^2 \) at the point \((x_1, y_1)\) is given by:
\[ xx_1 + yy_1 = r^2 \]
Here, \( r^2 = 50 \), \( x_1 = -7 \), and \( y_1 = 1 \).

Substituting these values into the formula gives:
\[ x(-7) + y(1) = 50 \]
This simplifies to:
\[ -7x + y = 50 \]

Rearranging gives the equation of the tangent line:
\[ 7x - y = -50 \]

### Step 3: Find the tangent line at the second intersection point.
Now, we will find the tangent line at the second intersection point \((-7, -1)\). Using the same formula:
\[ xx_1 + yy_1 = r^2 \]
Substituting \( x_1 = -7 \) and \( y_1 = -1 \):
\[ x(-7) + y(-1) = 50 \]
This simplifies to:
\[ -7x - y = 50 \]
Rearranging gives the equation of the tangent line:
\[ 7x + y = -50 \]

### Final Result
Thus, the equations of the tangents to the circle at the points of intersection are:
1. \( 7x - y = -50 \) (at the point \((-7, 1)\))
2. \( 7x + y = -50 \) (at the point \((-7, -1)\))