If the pole of a straight line with respect to the circle `x^(2)+y^(2)=a^(2)` lies on the circle `x^(2)+y^(2)=9a^(2)`, then the straight line touches the circle

Options

A. `9a^(2)=r^(2)`
B. `9r^(2)=a^(2)`
C. `r^(2)=a^(2)`
D. none of these

Answer

To solve the problem, we need to show that if the pole of a straight line with respect to the circle \(x^2 + y^2 = a^2\) lies on the circle \(x^2 + y^2 = 9a^2\), then the straight line touches the circle \(x^2 + y^2 = a^2\).

### Step-by-Step Solution:

1. **Understanding the Circles**:
- The first circle is given by the equation \(x^2 + y^2 = a^2\), which has a center at the origin (0,0) and a radius \(r_1 = a\).
- The second circle is given by the equation \(x^2 + y^2 = 9a^2\), which also has a center at the origin and a radius \(r_2 = 3a\).

2. **Finding the Pole**:
- The pole of a line with respect to a circle can be represented in polar coordinates. We assume the pole is at the point \((3a \cos \theta, 3a \sin \theta)\) where \(\theta\) is some angle.

3. **Equation of the Polar Line**:
- The equation of the polar line with respect to the circle \(x^2 + y^2 = a^2\) can be expressed as:
\[ x x_1 + y y_1 = r^2 \]
where \((x_1, y_1)\) is the pole and \(r\) is the radius of the first circle, \(a\).
- Substituting the coordinates of the pole:
\[ x (3a \cos \theta) + y (3a \sin \theta) = a^2 \]
- Simplifying this gives:
\[ 3a (x \cos \theta + y \sin \theta) = a^2 \]
- Dividing both sides by \(3a\) (assuming \(a \neq 0\)):
\[ x \cos \theta + y \sin \theta = \frac{a}{3} \]

4. **Finding the Perpendicular Distance from the Origin**:
- The distance \(d\) from the origin (0,0) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
- For our line, \(A = \cos \theta\), \(B = \sin \theta\), and \(C = -\frac{a}{3}\). Thus, we have:
\[ d = \frac{|\cos \theta \cdot 0 + \sin \theta \cdot 0 - \frac{a}{3}|}{\sqrt{\cos^2 \theta + \sin^2 \theta}} = \frac{\frac{a}{3}}{1} = \frac{a}{3} \]

5. **Condition for Tangency**:
- For the line to touch the circle \(x^2 + y^2 = a^2\), the distance from the center of the circle (the origin) to the line must equal the radius of the circle \(a\).
- We found that \(d = \frac{a}{3}\). For the line to be tangent to the circle, we need:
\[ d = r_1 \implies \frac{a}{3} = a \]
- This is not true, hence we need to check the condition again.

6. **Conclusion**:
- The condition for tangency is satisfied if the distance \(d\) is equal to the radius of the circle. Since we derived \(d = \frac{a}{3}\), we must have \(d = a\) for tangency, which is not satisfied in this case. Therefore, the line does not touch the circle.