The number of real tangents that can be drawn from (2, 2) to the circle `x^(2)+y^(2)-6x-4y+3=0`, is

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To determine the number of real tangents that can be drawn from the point (2, 2) to the circle defined by the equation \(x^2 + y^2 - 6x - 4y + 3 = 0\), we will follow these steps:

### Step 1: Rewrite the Circle's Equation
First, we rewrite the given equation of the circle in standard form. The equation is:
\[ x^2 + y^2 - 6x - 4y + 3 = 0 \]
We can rearrange it to complete the square.

### Step 2: Complete the Square
For \(x\):
\[ x^2 - 6x \rightarrow (x - 3)^2 - 9 \]
For \(y\):
\[ y^2 - 4y \rightarrow (y - 2)^2 - 4 \]
Substituting these back into the equation gives:
\[ (x - 3)^2 - 9 + (y - 2)^2 - 4 + 3 = 0 \]
Simplifying this:
\[ (x - 3)^2 + (y - 2)^2 - 10 = 0 \implies (x - 3)^2 + (y - 2)^2 = 10 \]
This shows that the circle has a center at \((3, 2)\) and a radius of \(\sqrt{10}\).

### Step 3: Find the Position of the Point (2, 2)
Next, we need to determine the position of the point (2, 2) with respect to the circle. We will use the formula:
\[ s_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c \]
where \((x_1, y_1) = (2, 2)\), \(g = -3\), \(f = -2\), and \(c = 10\).

### Step 4: Substitute Values into the Formula
Substituting the values:
\[ s_1 = 2^2 + 2^2 + 2(-3)(2) + 2(-2)(2) + 10 \]
Calculating each term:
\[ s_1 = 4 + 4 - 12 - 8 + 10 \]
Combining these:
\[ s_1 = 8 - 20 + 10 = -2 \]

### Step 5: Determine the Position of the Point
Since \(s_1 < 0\), this indicates that the point (2, 2) is inside the circle.

### Step 6: Conclusion
If the point is inside the circle, then no tangents can be drawn from that point to the circle. Therefore, the number of real tangents that can be drawn from the point (2, 2) to the circle is:
\[ \text{Answer: } 0 \]