The points of intersection of the two ellipses `x^2+2y^2-6x-12y + 23 = 0 and 4x^2 + 2y^2-20x-12y + 35 = 0`

Options

A. lie on a circle centred at `(8"/"3, 3)` and of radius `(1)/(3)sqrt((47)/(2))`
B. lie on a circle centred at `(–8"/"3, 3)` and of radius `(1)/(3)sqrt((47)/(2))`
C. lie on a circle centred at (8, 9) and of radius `(1)/(3)sqrt((47)/(2))`
D. are not cyclic

Answer

To find the points of intersection of the two ellipses given by the equations:

1. \( x^2 + 2y^2 - 6x - 12y + 23 = 0 \) (Equation 1)
2. \( 4x^2 + 2y^2 - 20x - 12y + 35 = 0 \) (Equation 2)

we can follow these steps:

### Step 1: Rearranging the equations
First, we can rearrange both equations to express them in a standard form.

**Equation 1:**
\[ x^2 - 6x + 2y^2 - 12y + 23 = 0 \]
Completing the square for \(x\) and \(y\):

- For \(x\):
\[ x^2 - 6x = (x - 3)^2 - 9 \]

- For \(y\):
\[ 2y^2 - 12y = 2(y^2 - 6y) = 2((y - 3)^2 - 9) = 2(y - 3)^2 - 18 \]

Substituting back:
\[ (x - 3)^2 - 9 + 2(y - 3)^2 - 18 + 23 = 0 \]
This simplifies to:
\[ (x - 3)^2 + 2(y - 3)^2 - 4 = 0 \]
or
\[ (x - 3)^2 + 2(y - 3)^2 = 4 \]

**Equation 2:**
\[ 4x^2 - 20x + 2y^2 - 12y + 35 = 0 \]
Completing the square for \(x\) and \(y\):

- For \(x\):
\[ 4(x^2 - 5x) = 4((x - \frac{5}{2})^2 - \frac{25}{4}) = 4(x - \frac{5}{2})^2 - 25 \]

- For \(y\):
\[ 2y^2 - 12y = 2(y^2 - 6y) = 2((y - 3)^2 - 9) = 2(y - 3)^2 - 18 \]

Substituting back:
\[ 4(x - \frac{5}{2})^2 - 25 + 2(y - 3)^2 - 18 + 35 = 0 \]
This simplifies to:
\[ 4(x - \frac{5}{2})^2 + 2(y - 3)^2 - 8 = 0 \]
or
\[ 4(x - \frac{5}{2})^2 + 2(y - 3)^2 = 8 \]
Dividing through by 8 gives:
\[ \frac{(x - \frac{5}{2})^2}{2} + \frac{(y - 3)^2}{4} = 1 \]

### Step 2: Setting the equations equal
Now we have two equations:
1. \((x - 3)^2 + 2(y - 3)^2 = 4\)
2. \(\frac{(x - \frac{5}{2})^2}{2} + \frac{(y - 3)^2}{4} = 1\)

We can express \(y\) in terms of \(x\) from one equation and substitute it into the other.

From the first equation:
\[ 2(y - 3)^2 = 4 - (x - 3)^2 \implies (y - 3)^2 = 2 - \frac{(x - 3)^2}{2} \]
Taking square roots:
\[ y - 3 = \pm \sqrt{2 - \frac{(x - 3)^2}{2}} \implies y = 3 \pm \sqrt{2 - \frac{(x - 3)^2}{2}} \]

### Step 3: Substituting into the second equation
Substituting \(y\) into the second equation will yield a quadratic in \(x\).

### Step 4: Solving for \(x\)
After substituting and simplifying, we will get a quadratic equation in \(x\). Solving this quadratic will give us the \(x\)-coordinates of the intersection points.

### Step 5: Finding corresponding \(y\) values
Substituting the \(x\) values back into either equation will give us the corresponding \(y\) values.

### Step 6: Final points of intersection
The final points of intersection will be the pairs \((x, y)\) that we calculated.