`(x-2)^2+y^2=36`
`y=-x+2`
The equation above represent a circle and a line that intersects the circle across its diameter . What is the point of intersection of the two equations that lies in Quadrant II?

Options

A. `(-3sqrt2,3sqrt2)`
B. `(-4,2)`
C. `(2+3,sqrt2)`
D. `(2-3sqrt2,3sqrt2)`

Answer

To find the point of intersection of the equations \((x-2)^2 + y^2 = 36\) and \(y = -x + 2\) that lies in Quadrant II, we can follow these steps:

### Step 1: Substitute the line equation into the circle equation
We start by substituting \(y = -x + 2\) into the circle's equation \((x-2)^2 + y^2 = 36\).

\[ (x-2)^2 + (-x + 2)^2 = 36 \]

### Step 2: Expand the equation
Now we expand both squares:

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

Combining these, we have:

\[ x^2 - 4x + 4 + x^2 - 4x + 4 = 36 \]

### Step 3: Combine like terms
Combine the terms on the left side:

\[ 2x^2 - 8x + 8 = 36 \]

### Step 4: Move all terms to one side
Now, we move 36 to the left side:

\[ 2x^2 - 8x + 8 - 36 = 0 \]
\[ 2x^2 - 8x - 28 = 0 \]

### Step 5: Simplify the equation
We can divide the entire equation by 2:

\[ x^2 - 4x - 14 = 0 \]

### Step 6: Use the quadratic formula
Now we apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -4\), and \(c = -14\):

\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-14)}}{2 \cdot 1} \]
\[ x = \frac{4 \pm \sqrt{16 + 56}}{2} \]
\[ x = \frac{4 \pm \sqrt{72}}{2} \]

### Step 7: Simplify the square root
Since \(\sqrt{72} = 6\sqrt{2}\):

\[ x = \frac{4 \pm 6\sqrt{2}}{2} \]
\[ x = 2 \pm 3\sqrt{2} \]

### Step 8: Determine the x-coordinate in Quadrant II
In Quadrant II, \(x\) must be negative. Therefore, we take:

\[ x = 2 - 3\sqrt{2} \]

### Step 9: Find the corresponding y-coordinate
Now we substitute \(x\) back into the line equation \(y = -x + 2\):

\[ y = -(2 - 3\sqrt{2}) + 2 \]
\[ y = -2 + 3\sqrt{2} + 2 \]
\[ y = 3\sqrt{2} \]

### Step 10: Write the point of intersection
Thus, the point of intersection in Quadrant II is:

\[ (2 - 3\sqrt{2}, 3\sqrt{2}) \]

### Final Answer
The point of intersection of the two equations that lies in Quadrant II is:

\[ \boxed{(2 - 3\sqrt{2}, 3\sqrt{2})} \]