If the line x+3y=0 is tangent at (0,0) to the circle of radius 1, then the centre of one such circle is

Options

A. `(3,0)`
B. `((-1)/(sqrt(10)),3/(sqrt(10)))`
C. `(3/(sqrt(10)),(-3)/(sqrt(10)))`
D. `(1/(sqrt(10)),3/(sqrt(10)))`

Answer

To solve the problem, we need to find the center of the circle given that the line \( x + 3y = 0 \) is tangent to the circle at the point \( (0, 0) \) and that the radius of the circle is 1.

### Step-by-Step Solution:

1. **Identify the Circle Equation**:
The general equation of a circle with center \( (h, k) \) and radius \( r \) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Since the radius is 1, we can write:
\[ (x - h)^2 + (y - k)^2 = 1 \]

2. **Point of Tangency**:
The point \( (0, 0) \) lies on the circle, so substituting \( (0, 0) \) into the circle equation gives:
\[ (0 - h)^2 + (0 - k)^2 = 1 \]
Simplifying this, we have:
\[ h^2 + k^2 = 1 \quad \text{(Equation 1)} \]

3. **Slope of the Tangent Line**:
The line \( x + 3y = 0 \) can be rewritten in slope-intercept form as:
\[ y = -\frac{1}{3}x \]
The slope of this tangent line is \( -\frac{1}{3} \).

4. **Slope of the Normal**:
The slope of the normal to the circle at the point of tangency is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal is:
\[ 3 \]

5. **Equation of the Normal**:
The normal line passes through the point \( (0, 0) \) and has a slope of \( 3 \). The equation of the normal line can be written as:
\[ y - 0 = 3(x - 0) \]
Simplifying this gives:
\[ y = 3x \quad \text{(Equation 2)} \]

6. **Finding the Center**:
Since the center \( (h, k) \) lies on the normal line, we can substitute \( k = 3h \) into Equation 1:
\[ h^2 + (3h)^2 = 1 \]
This simplifies to:
\[ h^2 + 9h^2 = 1 \]
\[ 10h^2 = 1 \]
\[ h^2 = \frac{1}{10} \]
Taking the square root gives:
\[ h = \pm \frac{1}{\sqrt{10}} \]

7. **Finding k**:
Now substituting \( h \) back into \( k = 3h \):
\[ k = 3 \left( \pm \frac{1}{\sqrt{10}} \right) = \pm \frac{3}{\sqrt{10}} \]

8. **Possible Centers**:
Therefore, the center of the circle can be:
\[ \left( \frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}} \right) \quad \text{or} \quad \left( -\frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}} \right) \]

### Final Answer:
The centers of the circle are:
\[ \left( \frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}} \right) \quad \text{and} \quad \left( -\frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}} \right) \]