The equation of the tangents to the circle `x^(2)+y^(2)=25` with slope 2 is

Options

A. `y=2x+-sqrt(5)`
B. `y=2x+-2sqrt(3)`
C. `y=2x+-3sqrt(5)`
D. `y=2x+-5sqrt(5)`

Answer

To find the equations of the tangents to the circle given by the equation \(x^2 + y^2 = 25\) with a slope of 2, we can follow these steps:

### Step 1: Identify the Circle's Parameters
The equation of the circle is \(x^2 + y^2 = 25\). This can be rewritten in standard form as:
\[ x^2 + y^2 = r^2 \]
where \(r^2 = 25\). Therefore, the radius \(r\) is:
\[ r = \sqrt{25} = 5 \]

### Step 2: Use the Tangent Equation
The general equation of a tangent to a circle \(x^2 + y^2 = r^2\) with slope \(m\) is given by:
\[ y = mx + c \]
In this case, the slope \(m\) is given as 2. Thus, we can write:
\[ y = 2x + c \]

### Step 3: Find the Value of \(c\)
To find \(c\), we can use the condition that the distance from the center of the circle (which is at the origin (0,0)) to the line must equal the radius of the circle. The distance \(d\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line \(y = 2x + c\), we can rewrite it in the form \(Ax + By + C = 0\):
\[ 2x - y + c = 0 \]
Here, \(A = 2\), \(B = -1\), and \(C = c\). The distance from the center (0,0) to the line is:
\[ d = \frac{|2(0) - 1(0) + c|}{\sqrt{2^2 + (-1)^2}} = \frac{|c|}{\sqrt{4 + 1}} = \frac{|c|}{\sqrt{5}} \]
Setting this distance equal to the radius \(r = 5\):
\[ \frac{|c|}{\sqrt{5}} = 5 \]

### Step 4: Solve for \(c\)
Multiplying both sides by \(\sqrt{5}\):
\[ |c| = 5\sqrt{5} \]
This gives us two possible values for \(c\):
\[ c = 5\sqrt{5} \quad \text{or} \quad c = -5\sqrt{5} \]

### Step 5: Write the Equations of the Tangents
Substituting the values of \(c\) back into the tangent equation \(y = 2x + c\):
1. For \(c = 5\sqrt{5}\):
\[ y = 2x + 5\sqrt{5} \]
2. For \(c = -5\sqrt{5}\):
\[ y = 2x - 5\sqrt{5} \]

### Final Answer
The equations of the tangents to the circle \(x^2 + y^2 = 25\) with slope 2 are:
\[ y = 2x + 5\sqrt{5} \quad \text{and} \quad y = 2x - 5\sqrt{5} \]