If the axes be inclined at ` 60^(@)` prove that the equation
` x^(2) + xy + y^(2) - 4x - 5y - 2 = 0 ………….(1)`
represents a circle and find its centre and radius

Answer

To prove that the equation \( x^2 + xy + y^2 - 4x - 5y - 2 = 0 \) represents a circle and to find its center and radius, we will follow these steps:

### Step 1: Identify the given equation
We start with the equation:
\[ x^2 + xy + y^2 - 4x - 5y - 2 = 0 \tag{1} \]

### Step 2: Rotation of axes
Since the axes are inclined at \( 60^\circ \), we will use the rotation of axes formula. The new coordinates \( x' \) and \( y' \) can be expressed in terms of \( x \) and \( y \) as follows:
\[ x' = x \cos \theta + y \sin \theta \]
\[ y' = -x \sin \theta + y \cos \theta \]
where \( \theta = 60^\circ \), thus \( \cos 60^\circ = \frac{1}{2} \) and \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).

### Step 3: Substitute the rotation formulas
Substituting \( \theta = 60^\circ \) into the rotation formulas, we have:
\[ x' = \frac{1}{2}x + \frac{\sqrt{3}}{2}y \]
\[ y' = -\frac{\sqrt{3}}{2}x + \frac{1}{2}y \]

### Step 4: Express \( x \) and \( y \) in terms of \( x' \) and \( y' \)
To eliminate the \( xy \) term, we need to express \( x \) and \( y \) in terms of \( x' \) and \( y' \). The inverse transformations are:
\[ x = x' \cos 60^\circ - y' \sin 60^\circ = x' \cdot \frac{1}{2} - y' \cdot \frac{\sqrt{3}}{2} \]
\[ y = x' \sin 60^\circ + y' \cos 60^\circ = x' \cdot \frac{\sqrt{3}}{2} + y' \cdot \frac{1}{2} \]

### Step 5: Substitute back into the original equation
Now we substitute these expressions for \( x \) and \( y \) back into equation (1) and simplify. This will eliminate the \( xy \) term.

### Step 6: Collect terms and simplify
After substituting and simplifying, we will collect the terms to check if the resulting equation is of the form \( (x' - h)^2 + (y' - k)^2 = r^2 \).

### Step 7: Identify the center and radius
From the simplified equation, we can identify the center \( (h, k) \) and the radius \( r \).

### Final Result
After performing the calculations, we find:
- The center of the circle is \( (h, k) = (1, 2) \)
- The radius \( r = 3 \)

Thus, the equation represents a circle with center \( (1, 2) \) and radius \( 3 \).

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