Equation of tangents drawn from `(0, 0)` to `x^(2) + y^(2) - 6x -6y + 9 = 0` are

Options

A. x = y
B. y = 0
C. x = 0
D. `x + y = 0`

Answer

To find the equations of the tangents drawn from the point (0, 0) to the circle given by the equation \(x^2 + y^2 - 6x - 6y + 9 = 0\), we can follow these steps:

### Step 1: Rewrite the Circle's Equation
First, we rewrite the equation of the circle in standard form. The given equation is:

\[ x^2 + y^2 - 6x - 6y + 9 = 0 \]

We can complete the square for both \(x\) and \(y\):

\[ (x^2 - 6x) + (y^2 - 6y) + 9 = 0 \]

Completing the square:

\[ (x - 3)^2 - 9 + (y - 3)^2 - 9 + 9 = 0 \]

This simplifies to:

\[ (x - 3)^2 + (y - 3)^2 = 9 \]

This represents a circle with center at \((3, 3)\) and radius \(3\).

### Step 2: Equation of the Tangent Line
The equation of any line passing through the origin (0, 0) can be expressed as:

\[ y = mx \]

where \(m\) is the slope of the line.

### Step 3: Substitute the Line Equation into the Circle's Equation
Substituting \(y = mx\) into the circle's equation:

\[ (x - 3)^2 + (mx - 3)^2 = 9 \]

Expanding this:

\[ (x - 3)^2 + (m^2x^2 - 6mx + 9) = 9 \]

This simplifies to:

\[ (x - 3)^2 + m^2x^2 - 6mx = 0 \]

Expanding \((x - 3)^2\):

\[ x^2 - 6x + 9 + m^2x^2 - 6mx = 0 \]

Combining like terms:

\[ (1 + m^2)x^2 + (-6 - 6m)x + 9 = 0 \]

### Step 4: Condition for Tangency
For the line to be tangent to the circle, the discriminant of this quadratic equation must be zero:

\[ b^2 - 4ac = 0 \]

Here, \(a = 1 + m^2\), \(b = -6 - 6m\), and \(c = 9\).

Calculating the discriminant:

\[ (-6 - 6m)^2 - 4(1 + m^2)(9) = 0 \]

Expanding this:

\[ 36 + 72m + 36m^2 - 36 - 36m^2 = 0 \]

This simplifies to:

\[ 72m = 0 \]

Thus, we find:

\[ m = 0 \]

### Step 5: Finding the Tangent Lines
The first tangent line corresponds to \(m = 0\):

\[ y = 0 \]

This is the x-axis.

### Step 6: Considering the Other Tangent
Since the point (0, 0) lies outside the circle, there will be another tangent line. The cancellation of \(m^2\) indicates that there is another slope, which is infinite. This corresponds to a vertical line:

\[ x = 0 \]

### Final Answer
The equations of the tangents drawn from the point (0, 0) to the circle are:

1. \(y = 0\) (the x-axis)
2. \(x = 0\) (the y-axis)