Step 1: Understanding the Question:
We need to find the equations of the tangent lines to a given circle. The condition is that these tangents must be perpendicular to a given line.
Step 2: Key Formula or Approach:
1. Find the slope of the given line.
2. Find the slope of the required tangents using the condition of perpendicularity (\(m_1 m_2 = -1\)).
3. Use the equation of a tangent to a circle \(x^2 + y^2 = r^2\) with slope \(m\): \(y = mx \pm r\sqrt{1+m^2}\).
4. Rearrange the equation into the form given in the options.
Step 3: Detailed Explanation:
1. Analyze the circle:
The equation of the circle is \(x^2 + y^2 = 36\).
This is a circle centered at the origin (0,0) with radius \(r^2 = 36\), so \(r=6\).
2. Find the slope of the tangent:
The given line is \(5x + y - 2 = 0\). We can write it in slope-intercept form \(y = -5x + 2\).
The slope of this line is \(m_1 = -5\).
The tangents are perpendicular to this line. Let the slope of the tangent be \(m\).
The condition for perpendicular lines is \(m \times m_1 = -1\).
\[ m \times (-5) = -1 \implies m = \frac{-1}{-5} = \frac{1}{5} \]
So, the slope of the required tangents is \(m = 1/5\).
3. Find the equation of the tangents:
The equation of a tangent with slope \(m\) to the circle \(x^2+y^2=r^2\) is:
\[ y = mx \pm r\sqrt{1+m^2} \]
Substitute \(m = 1/5\) and \(r = 6\):
\[ y = \frac{1}{5}x \pm 6\sqrt{1 + \left(\frac{1}{5}\right)^2} \]
\[ y = \frac{1}{5}x \pm 6\sqrt{1 + \frac{1}{25}} \]
\[ y = \frac{1}{5}x \pm 6\sqrt{\frac{25+1}{25}} \]
\[ y = \frac{1}{5}x \pm 6\sqrt{\frac{26}{25}} \]
\[ y = \frac{1}{5}x \pm 6\frac{\sqrt{26}}{5} \]
4. Rearrange the equation:
Multiply the entire equation by 5 to clear the denominator:
\[ 5y = x \pm 6\sqrt{26} \]
Rearrange to match the form in the options:
\[ x - 5y \pm 6\sqrt{26} = 0 \]
This equation represents the two parallel tangents.
Step 4: Final Answer:
The equations of the tangents are \( x - 5y \pm 6\sqrt{26} = 0 \), which matches option (A).