Find the point to which the origin has to be shifted to eliminate x and y terms (to remove first degree terms) in the following equations.
`x^(2)+y^(2)-2ax-4ay+a^(2) = 0`

Answer

To find the point to which the origin has to be shifted to eliminate the first degree terms in the equation \( x^2 + y^2 - 2ax - 4ay + a^2 = 0 \), we can follow these steps:

### Step 1: Identify the original equation
We start with the given equation:
\[ x^2 + y^2 - 2ax - 4ay + a^2 = 0 \]

### Step 2: Shift the origin
Let’s denote the new origin as \( (H, K) \). We will express the old coordinates \( (x, y) \) in terms of the new coordinates \( (X, Y) \):
\[ x = X + H \quad \text{and} \quad y = Y + K \]

### Step 3: Substitute the new coordinates into the equation
Substituting \( x \) and \( y \) into the original equation gives:
\[ (X + H)^2 + (Y + K)^2 - 2a(X + H) - 4a(Y + K) + a^2 = 0 \]

### Step 4: Expand the equation
Now we expand the equation:
\[ (X^2 + 2HX + H^2) + (Y^2 + 2KY + K^2) - 2aX - 2aH - 4aY - 4aK + a^2 = 0 \]

Combining like terms, we have:
\[ X^2 + Y^2 + (2H - 2a)X + (2K - 4a)Y + (H^2 + K^2 - 2aH - 4aK + a^2) = 0 \]

### Step 5: Eliminate the first degree terms
To eliminate the first degree terms, we set the coefficients of \( X \) and \( Y \) to zero:
1. \( 2H - 2a = 0 \)
2. \( 2K - 4a = 0 \)

### Step 6: Solve for \( H \) and \( K \)
From the first equation:
\[ 2H - 2a = 0 \implies H = a \]

From the second equation:
\[ 2K - 4a = 0 \implies K = 2a \]

### Step 7: Write the new origin
Thus, the new origin to which the origin has to be shifted is:
\[ (H, K) = (a, 2a) \]

### Final Answer
The point to which the origin has to be shifted is \( (a, 2a) \).

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