If one end of a diameter of the circle `2x^(2)+2y^(2)-4x-8y+2=0` is (-1,2), then the other end of the diameter is

Options

A. (2,1)
B. (3,2)
C. (4,3)
D. (5,4)

Answer

To find the other end of the diameter of the circle given the equation \(2x^2 + 2y^2 - 4x - 8y + 2 = 0\) and one end of the diameter at the point \((-1, 2)\), we can follow these steps:

### Step 1: Rewrite the equation of the circle in standard form
First, we simplify the equation by dividing everything by 2:

\[ x^2 + y^2 - 2x - 4y + 1 = 0 \]

### Step 2: Rearranging the equation
Next, we rearrange the equation to group the \(x\) and \(y\) terms:

\[ x^2 - 2x + y^2 - 4y + 1 = 0 \]

### Step 3: Completing the square
Now, we complete the square for the \(x\) and \(y\) terms.

For \(x^2 - 2x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]

For \(y^2 - 4y\):
\[ y^2 - 4y = (y - 2)^2 - 4 \]

Substituting these back into the equation gives:

\[ (x - 1)^2 - 1 + (y - 2)^2 - 4 + 1 = 0 \]

This simplifies to:

\[ (x - 1)^2 + (y - 2)^2 - 4 = 0 \]

Thus, we have:

\[ (x - 1)^2 + (y - 2)^2 = 4 \]

### Step 4: Identify the center and radius
The center of the circle is \((1, 2)\) and the radius is \(2\) (since \(4\) is \(2^2\)).

### Step 5: Finding the other end of the diameter
Let the other end of the diameter be \((x, y)\). The midpoint of the diameter is the center of the circle. Therefore, we can set up the following equations based on the midpoint formula:

\[ \left(\frac{-1 + x}{2}, \frac{2 + y}{2}\right) = (1, 2) \]

From the x-coordinates:

\[ \frac{-1 + x}{2} = 1 \]
Multiplying both sides by 2:

\[ -1 + x = 2 \implies x = 3 \]

From the y-coordinates:

\[ \frac{2 + y}{2} = 2 \]
Multiplying both sides by 2:

\[ 2 + y = 4 \implies y = 2 \]

### Conclusion
Thus, the coordinates of the other end of the diameter are \((3, 2)\).