Let `vec(r)` is a positive vector of a variable pont in cartesian OXY plane such that `vecr.(10hatj-8hati-vecr)=40` and `p_1=max{|vecr+2hati-3hatj|^2},p_2=min{|vecr+2hati-3hatj|^2}`. Then `p_1 + p_2` ​is equal to

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the given problem step-by-step, we will follow the instructions provided in the video transcript and derive the necessary calculations.

### Step 1: Understand the given equation
We start with the equation:
\[ \vec{r} \cdot (10 \hat{j} - 8 \hat{i} - \vec{r}) = 40 \]
This can be rewritten as:
\[ \vec{r} \cdot (10 \hat{j} - 8 \hat{i}) - \vec{r} \cdot \vec{r} = 40 \]

### Step 2: Express \(\vec{r}\) in terms of its components
Let \(\vec{r} = x \hat{i} + y \hat{j}\). Then, we can substitute this into the equation:
\[ (x \hat{i} + y \hat{j}) \cdot (10 \hat{j} - 8 \hat{i}) - (x^2 + y^2) = 40 \]
Calculating the dot product:
\[ -8x + 10y - (x^2 + y^2) = 40 \]

### Step 3: Rearranging the equation
Rearranging gives:
\[ x^2 + y^2 + 8x - 10y + 40 = 0 \]

### Step 4: Completing the square
To find the center and radius of the circle, we complete the square:
\[ (x^2 + 8x) + (y^2 - 10y) + 40 = 0 \]
Completing the square:
\[ (x + 4)^2 - 16 + (y - 5)^2 - 25 + 40 = 0 \]
This simplifies to:
\[ (x + 4)^2 + (y - 5)^2 - 1 = 0 \]
Which implies:
\[ (x + 4)^2 + (y - 5)^2 = 1 \]
This is a circle with center at \((-4, 5)\) and radius \(1\).

### Step 5: Finding \(p_1\)
We need to find:
\[ p_1 = \max \{ |\vec{r} + 2 \hat{i} - 3 \hat{j}|^2 \} \]
This is the distance from the point \((-2, 3)\) to the circle. The maximum distance is the distance from \((-2, 3)\) to the center \((-4, 5)\) plus the radius:
\[ \text{Distance} = \sqrt{((-2) - (-4))^2 + ((3) - (5))^2} + 1 \]
Calculating:
\[ = \sqrt{(2)^2 + (2)^2} + 1 = \sqrt{4 + 4} + 1 = 2\sqrt{2} + 1 \]
Thus:
\[ p_1 = (2\sqrt{2} + 1)^2 = 4 \cdot 2 + 4 \cdot 2 + 1 = 8 + 4\sqrt{2} + 1 = 9 + 4\sqrt{2} \]

### Step 6: Finding \(p_2\)
Now, we find:
\[ p_2 = \min \{ |\vec{r} + 2 \hat{i} - 3 \hat{j}|^2 \} \]
This is the minimum distance from the point \((-2, 3)\) to the circle. The minimum distance is the distance from \((-2, 3)\) to the center \((-4, 5)\) minus the radius:
\[ \text{Distance} = \sqrt{((-2) - (-4))^2 + ((3) - (5))^2} - 1 \]
Calculating:
\[ = \sqrt{(2)^2 + (2)^2} - 1 = \sqrt{4 + 4} - 1 = 2\sqrt{2} - 1 \]
Thus:
\[ p_2 = (2\sqrt{2} - 1)^2 = 4 \cdot 2 - 4\cdot 2 + 1 = 8 - 4\sqrt{2} + 1 = 9 - 4\sqrt{2} \]

### Step 7: Finding \(p_1 + p_2\)
Now, we add \(p_1\) and \(p_2\):
\[ p_1 + p_2 = (9 + 4\sqrt{2}) + (9 - 4\sqrt{2}) = 18 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{18} \]