How many tangents can be drawn from the poin (3,-2) to the circle `x^(2)+y^(2)-8x-6y+9=0`?

Options

A. 2
B. 1
C. 0
D. None of these

Answer

To determine how many tangents can be drawn from the point (3, -2) to the circle given by the equation \(x^2 + y^2 - 8x - 6y + 9 = 0\), we will follow these steps:

### Step 1: Rewrite the Circle's Equation in Standard Form
The given equation of the circle is:
\[ x^2 + y^2 - 8x - 6y + 9 = 0 \]
We can rewrite this equation in standard form by completing the square for \(x\) and \(y\).

1. Group the \(x\) and \(y\) terms:
\[ (x^2 - 8x) + (y^2 - 6y) + 9 = 0 \]

2. Complete the square:
- For \(x^2 - 8x\):
\[ x^2 - 8x = (x - 4)^2 - 16 \]
- For \(y^2 - 6y\):
\[ y^2 - 6y = (y - 3)^2 - 9 \]

3. Substitute back into the equation:
\[ (x - 4)^2 - 16 + (y - 3)^2 - 9 + 9 = 0 \]
Simplifying gives:
\[ (x - 4)^2 + (y - 3)^2 - 16 = 0 \]
Thus, we have:
\[ (x - 4)^2 + (y - 3)^2 = 16 \]
This represents a circle with center \((4, 3)\) and radius \(4\).

### Step 2: Calculate the Distance from the Point to the Center of the Circle
Next, we find the distance \(d\) from the point \((3, -2)\) to the center of the circle \((4, 3)\) using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{(4 - 3)^2 + (3 - (-2))^2} \]
Calculating this gives:
\[ d = \sqrt{(1)^2 + (3 + 2)^2} = \sqrt{1 + 5^2} = \sqrt{1 + 25} = \sqrt{26} \]

### Step 3: Compare the Distance to the Radius
Now, we compare the distance \(d\) to the radius \(r\) of the circle:
- Radius \(r = 4\)
- Distance \(d = \sqrt{26} \approx 5.1\)

Since \(d > r\), the point \((3, -2)\) lies outside the circle.

### Step 4: Determine the Number of Tangents
If a point lies outside a circle, two tangents can be drawn from that point to the circle. Thus, the number of tangents that can be drawn from the point \((3, -2)\) to the circle is:

\[ \text{Number of tangents} = 2 \]

### Final Answer
The answer is that **2 tangents can be drawn** from the point (3, -2) to the circle.

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