Find the locus of the point which is at a constant distance of 5 units from (4,-3).

Answer

To find the locus of the point that is at a constant distance of 5 units from the point (4, -3), we can follow these steps:

### Step 1: Define the point and distance
Let the point be \( P(x, y) \). The distance from point \( P \) to the point \( (4, -3) \) is given by the distance formula:
\[ d = \sqrt{(x - 4)^2 + (y + 3)^2} \]
We know that this distance \( d \) is equal to 5 units.

### Step 2: Set up the equation
We can set up the equation based on the distance:
\[ \sqrt{(x - 4)^2 + (y + 3)^2} = 5 \]

### Step 3: Square both sides
To eliminate the square root, we square both sides of the equation:
\[ (x - 4)^2 + (y + 3)^2 = 5^2 \]
This simplifies to:
\[ (x - 4)^2 + (y + 3)^2 = 25 \]

### Step 4: Expand the equation
Now, we will expand the left-hand side:
\[ (x - 4)^2 = x^2 - 8x + 16 \]
\[ (y + 3)^2 = y^2 + 6y + 9 \]
Combining these, we have:
\[ x^2 - 8x + 16 + y^2 + 6y + 9 = 25 \]

### Step 5: Combine like terms
Now, we combine the constant terms:
\[ x^2 + y^2 - 8x + 6y + 25 = 25 \]
Subtracting 25 from both sides gives:
\[ x^2 + y^2 - 8x + 6y = 0 \]

### Step 6: Write the final equation
This is the equation of the locus of the point that is at a constant distance of 5 units from the point (4, -3):
\[ x^2 + y^2 - 8x + 6y = 0 \]