Find the equation of the tangent to the circle `x^2 + y^2 = 9`, which passes through the point on the circle where x = 2 and y is positive.

Options

A. `sqrt5y = 2x -9`
B. `sqrt5y = -2x +9`
C. `2y=-sqrt5x + 4sqrt5`
D. `2y=-sqrt5x - 4sqrt5`

Answer

To find the equation of the tangent to the circle \(x^2 + y^2 = 9\) that passes through the point on the circle where \(x = 2\) and \(y\) is positive, we can follow these steps:

### Step 1: Find the coordinates of the point on the circle
We start with the equation of the circle:

\[ x^2 + y^2 = 9 \]

Substituting \(x = 2\):

\[ 2^2 + y^2 = 9 \]

This simplifies to:

\[ 4 + y^2 = 9 \]

Subtracting 4 from both sides gives:

\[ y^2 = 5 \]

Taking the positive square root (since \(y\) is positive):

\[ y = \sqrt{5} \]

So, the point on the circle is \((2, \sqrt{5})\).

### Step 2: Determine the slope of the radius at the point
The center of the circle is at the origin \((0, 0)\). The slope \(m\) of the radius from the center to the point \((2, \sqrt{5})\) is given by:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\sqrt{5} - 0}{2 - 0} = \frac{\sqrt{5}}{2} \]

### Step 3: Find the slope of the tangent line
The slope of the tangent line is the negative reciprocal of the slope of the radius. Therefore, the slope \(m_t\) of the tangent line is:

\[ m_t = -\frac{1}{m} = -\frac{2}{\sqrt{5}} \]

### Step 4: Use the point-slope form of the equation of a line
The equation of the tangent line can be written using the point-slope form:

\[ y - y_1 = m_t(x - x_1) \]

Substituting \(y_1 = \sqrt{5}\), \(m_t = -\frac{2}{\sqrt{5}}\), and \(x_1 = 2\):

\[ y - \sqrt{5} = -\frac{2}{\sqrt{5}}(x - 2) \]

### Step 5: Simplify the equation
Distributing the slope on the right side:

\[ y - \sqrt{5} = -\frac{2}{\sqrt{5}}x + \frac{4}{\sqrt{5}} \]

Adding \(\sqrt{5}\) to both sides:

\[ y = -\frac{2}{\sqrt{5}}x + \frac{4}{\sqrt{5}} + \sqrt{5} \]

To combine the terms on the right, convert \(\sqrt{5}\) into a fraction with a common denominator:

\[ \sqrt{5} = \frac{5}{\sqrt{5}} \]

Thus:

\[ y = -\frac{2}{\sqrt{5}}x + \frac{4 + 5}{\sqrt{5}} = -\frac{2}{\sqrt{5}}x + \frac{9}{\sqrt{5}} \]

### Step 6: Rearranging to standard form
Multiplying through by \(\sqrt{5}\) to eliminate the fraction:

\[ \sqrt{5}y = -2x + 9 \]

Rearranging gives:

\[ 2x + \sqrt{5}y - 9 = 0 \]

This is the equation of the tangent line.

### Final Answer
The equation of the tangent to the circle \(x^2 + y^2 = 9\) that passes through the point \((2, \sqrt{5})\) is:

\[ 2x + \sqrt{5}y - 9 = 0 \]