The equation of tangents drawn from the point (0,1) to the circle `x^2+y^2-4x-2y+4=0` are

Options

A. 2x-y+1=0 ,x+2y-2=0
B. 2x-y-1=0 ,x+2y-2=0
C. 2x-y+1=0,x+2y+2=0
D. `x=+-sqrt(3)(y-1)`

Answer

To find the equations of the tangents drawn from the point (0, 1) to the circle given by the equation \( x^2 + y^2 - 4x - 2y + 4 = 0 \), we will follow these steps:

### Step 1: Rewrite the Circle's Equation
First, we need to rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 - 4x - 2y + 4 = 0 \]
We can complete the square for the \(x\) and \(y\) terms.

For \(x^2 - 4x\):
\[ x^2 - 4x = (x - 2)^2 - 4 \]

For \(y^2 - 2y\):
\[ y^2 - 2y = (y - 1)^2 - 1 \]

Substituting these back into the equation gives:
\[ (x - 2)^2 - 4 + (y - 1)^2 - 1 + 4 = 0 \]
Simplifying this:
\[ (x - 2)^2 + (y - 1)^2 - 1 = 0 \]
\[ (x - 2)^2 + (y - 1)^2 = 1 \]
This shows that the circle has center \( (2, 1) \) and radius \( 1 \).

### Step 2: Use the Tangent Formula
The formula for the equation of the tangents from an external point \( (x_1, y_1) \) to the circle \( (x - h)^2 + (y - k)^2 = r^2 \) is given by:
\[ (x - h)(x_1 - h) + (y - k)(y_1 - k) = r^2 \]
Here, \( (h, k) = (2, 1) \), \( r = 1 \), and \( (x_1, y_1) = (0, 1) \).

### Step 3: Substitute the Values
Substituting the values into the tangent formula:
\[ (x - 2)(0 - 2) + (y - 1)(1 - 1) = 1^2 \]
This simplifies to:
\[ -2(x - 2) + 0 = 1 \]
\[ -2x + 4 = 1 \]
\[ -2x = 1 - 4 \]
\[ -2x = -3 \]
\[ x = \frac{3}{2} \]

### Step 4: Finding the Tangent Equations
Now, we need to find the equations of the tangents. We can use the point-slope form of the line. The slope of the tangent from the point \( (0, 1) \) to the circle at the point of tangency can be calculated using the center of the circle.

The slope \( m \) of the line connecting the center \( (2, 1) \) to the point \( (0, 1) \) is:
\[ m = \frac{1 - 1}{0 - 2} = 0 \]
This means the tangents will be vertical lines at \( x = \frac{3}{2} \).

### Step 5: Final Tangent Equations
The equations of the tangents from the point \( (0, 1) \) to the circle are:
\[ y - 1 = m(x - 0) \]
Since we have two tangents, we can find the two slopes using the distance formula or by finding the points of tangency.

After simplification, we find that the equations of the tangents are:
\[ y = 1 \pm \sqrt{3}(x - 0) \]
Thus, the final tangent equations are:
\[ y = \sqrt{3}x + 1 \quad \text{and} \quad y = -\sqrt{3}x + 1 \]