The locus of the point which is at a distance 5 unit from ` ( -2, 3 ) ` is

Answer

To find the locus of a point that is at a distance of 5 units from the point (-2, 3), we can follow these steps:

### Step 1: Define the point and distance
Let the point be represented as (x, y). The distance from this point to the point (-2, 3) is given as 5 units.

### Step 2: Use the distance formula
According to the distance formula, the distance \( d \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Here, we have:
- \((x_1, y_1) = (x, y)\)
- \((x_2, y_2) = (-2, 3)\)
- \(d = 5\)

Substituting these values into the distance formula gives us:
\[ \sqrt{(x - (-2))^2 + (y - 3)^2} = 5 \]
This simplifies to:
\[ \sqrt{(x + 2)^2 + (y - 3)^2} = 5 \]

### Step 3: Square both sides
To eliminate the square root, we square both sides of the equation:
\[ (x + 2)^2 + (y - 3)^2 = 5^2 \]
This simplifies to:
\[ (x + 2)^2 + (y - 3)^2 = 25 \]

### Step 4: Expand the equation
Now, we expand the left side:
\[ (x + 2)^2 = x^2 + 4x + 4 \]
\[ (y - 3)^2 = y^2 - 6y + 9 \]
Combining these, we have:
\[ x^2 + 4x + 4 + y^2 - 6y + 9 = 25 \]

### Step 5: Combine like terms
Now, combine all the terms:
\[ x^2 + y^2 + 4x - 6y + 13 = 25 \]
Subtracting 25 from both sides gives:
\[ x^2 + y^2 + 4x - 6y - 12 = 0 \]

### Final Result
The locus of the point that is at a distance of 5 units from the point (-2, 3) is given by the equation:
\[ x^2 + y^2 + 4x - 6y - 12 = 0 \]

---