The points of contact of tangents to the circle `x^(2)+y^(2)=25` which are inclined at an angle of `30^(@)` to the x-axis are

Options

A. `(pm 5//2, pm 1//2)`
B. `(pm1//2, pm5//2)`
C. `(pm 5//2, pm 1//2)`
D. none of these

Answer

To solve the problem of finding the points of contact of tangents to the circle \(x^2 + y^2 = 25\) that are inclined at an angle of \(30^\circ\) to the x-axis, we can follow these steps:

### Step 1: Identify the Circle
The given equation of the circle is:
\[ x^2 + y^2 = 25 \]
This can be compared to the general form of a circle centered at the origin, \(x^2 + y^2 = a^2\), where \(a\) is the radius. Here, \(a^2 = 25\), so:
\[ a = 5 \]

### Step 2: Determine the Slope of the Tangents
The angle of inclination of the tangents to the x-axis is given as \(30^\circ\). The slope \(m\) of the tangent can be calculated using:
\[ m = \tan(30^\circ) = \frac{1}{\sqrt{3}} \]
Since tangents can have both positive and negative slopes, we consider:
\[ m = \pm \frac{1}{\sqrt{3}} \]

### Step 3: Write the Equation of the Tangent
The equation of the tangent to the circle in slope form is given by:
\[ y = mx + \sqrt{a^2(1 + m^2)} \]
Substituting \(a = 5\) and \(m = \pm \frac{1}{\sqrt{3}}\):
\[ y = \pm \frac{1}{\sqrt{3}}x + \sqrt{25(1 + \left(\frac{1}{\sqrt{3}}\right)^2)} \]
Calculating \(1 + \left(\frac{1}{\sqrt{3}}\right)^2\):
\[ 1 + \frac{1}{3} = \frac{4}{3} \]
Thus:
\[ \sqrt{25 \cdot \frac{4}{3}} = 5 \cdot \frac{2}{\sqrt{3}} = \frac{10}{\sqrt{3}} \]
So the tangent equations become:
\[ y = \pm \frac{1}{\sqrt{3}}x + \frac{10}{\sqrt{3}} \]

### Step 4: General Form of Tangent Equation
The general form of the tangent to the circle can also be expressed as:
\[ x_1 x + y_1 y = a^2 \]
where \((x_1, y_1)\) is the point of contact. Rearranging gives:
\[ y = -\frac{x_1}{y_1}x + \frac{25}{y_1} \]

### Step 5: Compare Coefficients
From the tangent equations, we have:
\[ -\frac{x_1}{y_1} = \pm \frac{1}{\sqrt{3}} \quad \text{and} \quad \frac{25}{y_1} = \frac{10}{\sqrt{3}} \]

### Step 6: Solve for \(y_1\)
From the second equation:
\[ y_1 = \frac{25 \sqrt{3}}{10} = \frac{5 \sqrt{3}}{2} \]
Now substituting \(y_1\) back into the first equation:
\[ -\frac{x_1}{\frac{5 \sqrt{3}}{2}} = \pm \frac{1}{\sqrt{3}} \]
This simplifies to:
\[ x_1 = \mp \frac{5}{2} \]

### Step 7: Final Points of Contact
Thus, the points of contact are:
\[ \left(\frac{5}{2}, \frac{5 \sqrt{3}}{2}\right), \left(-\frac{5}{2}, \frac{5 \sqrt{3}}{2}\right), \left(\frac{5}{2}, -\frac{5 \sqrt{3}}{2}\right), \left(-\frac{5}{2}, -\frac{5 \sqrt{3}}{2}\right) \]

### Final Answer
The points of contact of the tangents to the circle are:
\[ \left(\frac{5}{2}, \frac{5\sqrt{3}}{2}\right), \left(-\frac{5}{2}, \frac{5\sqrt{3}}{2}\right), \left(\frac{5}{2}, -\frac{5\sqrt{3}}{2}\right), \left(-\frac{5}{2}, -\frac{5\sqrt{3}}{2}\right) \]