If line `3x + y=0` be a tangent to a circle drawn from origin to a circle centred at the point (2,- 1) then the equation of other tangent through the origin is

Options

A. `x-3y=0`
B. `x+2y=0`
C. `x+3y=0`
D. `3x-y=0`

Answer

To find the equation of the other tangent from the origin to the circle centered at (2, -1) given that the line \(3x + y = 0\) is a tangent, we can follow these steps:

### Step 1: Identify the center and radius of the circle
The center of the circle is given as \(C(2, -1)\). Since we have a tangent line, we need to find the radius of the circle.

### Step 2: Use the formula for the distance from a point to a line
The distance \(r\) from the center of the circle \(C(2, -1)\) to the tangent line \(3x + y = 0\) can be calculated using the formula:
\[ r = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
where \(Ax + By + C = 0\) is the equation of the line, and \((x_1, y_1)\) is the center of the circle.

Here, \(A = 3\), \(B = 1\), \(C = 0\), and \((x_1, y_1) = (2, -1)\).

### Step 3: Calculate the distance
Substituting the values into the formula:
\[ r = \frac{|3(2) + 1(-1) + 0|}{\sqrt{3^2 + 1^2}} = \frac{|6 - 1|}{\sqrt{9 + 1}} = \frac{5}{\sqrt{10}} = \frac{5\sqrt{10}}{10} = \frac{\sqrt{10}}{2} \]

### Step 4: Set up the equation of the other tangent
The equation of the tangent line from the origin can be expressed as \(y = mx\), where \(m\) is the slope. The distance from the center of the circle to this line must also equal the radius \(r\).

### Step 5: Use the distance formula again
The distance from the center \(C(2, -1)\) to the line \(y = mx\) is given by:
\[ \frac{|m(2) - 1|}{\sqrt{m^2 + 1}} = \frac{\sqrt{10}}{2} \]

### Step 6: Solve for \(m\)
Squaring both sides gives:
\[ \frac{(m(2) - 1)^2}{m^2 + 1} = \frac{10}{4} \]
\[ 4(m(2) - 1)^2 = 10(m^2 + 1) \]
Expanding and simplifying:
\[ 4(4m^2 - 4m + 1) = 10m^2 + 10 \]
\[ 16m^2 - 16m + 4 = 10m^2 + 10 \]
\[ 6m^2 - 16m - 6 = 0 \]

### Step 7: Factor or use the quadratic formula
Dividing through by 2:
\[ 3m^2 - 8m - 3 = 0 \]
Now we can factor or use the quadratic formula:
\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{8 \pm \sqrt{(-8)^2 - 4(3)(-3)}}{2(3)} = \frac{8 \pm \sqrt{64 + 36}}{6} = \frac{8 \pm 10}{6} \]
This gives us:
\[ m_1 = 3 \quad \text{and} \quad m_2 = -\frac{1}{3} \]

### Step 8: Write the equations of the tangents
Using the slopes:
1. For \(m_1 = 3\): \(y = 3x\) or \(3x - y = 0\)
2. For \(m_2 = -\frac{1}{3}\): \(y = -\frac{1}{3}x\) or \(x + 3y = 0\)

### Final Result
The equations of the tangents from the origin to the circle are:
1. \(3x - y = 0\)
2. \(x + 3y = 0\)