If `P and Q` are two points on the circle `x^2 + y^2 - 4x + 6y-3=0` which are farthest and nearest respectively from the point `(7, 2)` then.

(A) `P-=(2-2sqrt(2), -3-2sqrt(2))`
(B) `Q-= (2+2sqrt(2), -3 + 2 sqrt(2))`
(C) `P-= (2+2sqrt(2), -3 + 2 sqrt(2))`
(D) `Q-= (2-2sqrt(2), -3 + 2 sqrt(2))`

Options

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

Answer

To solve the problem, we need to find the points on the circle that are farthest and nearest to the point (7, 2). Here’s a step-by-step solution:

### Step 1: Rewrite the Circle Equation
The given equation of the circle is:
\[ x^2 + y^2 - 4x + 6y - 3 = 0 \]

We can complete the square for both \(x\) and \(y\).

1. For \(x\):
\[ x^2 - 4x = (x - 2)^2 - 4 \]

2. For \(y\):
\[ y^2 + 6y = (y + 3)^2 - 9 \]

Substituting back into the equation gives:
\[ (x - 2)^2 - 4 + (y + 3)^2 - 9 - 3 = 0 \]
\[ (x - 2)^2 + (y + 3)^2 - 16 = 0 \]
\[ (x - 2)^2 + (y + 3)^2 = 16 \]

### Step 2: Identify the Center and Radius
From the equation \((x - 2)^2 + (y + 3)^2 = 16\), we can identify:
- Center \(C = (2, -3)\)
- Radius \(r = 4\)

### Step 3: Calculate the Distance from the Center to the Point (7, 2)
Next, we calculate the distance from the center \(C(2, -3)\) to the point \(A(7, 2)\):
\[ d = \sqrt{(7 - 2)^2 + (2 + 3)^2} = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2} \]

### Step 4: Determine the Farthest and Nearest Points
To find the farthest and nearest points on the circle from point \(A(7, 2)\), we need to consider the line connecting \(C\) and \(A\) and extend it to the circle.

The distance from \(C\) to \(A\) is \(5\sqrt{2}\) and the radius is \(4\). The nearest point \(Q\) on the circle will be towards \(A\) and the farthest point \(P\) will be away from \(A\).

1. **Nearest Point \(Q\)**:
\[ d_{Q} = d - r = 5\sqrt{2} - 4 \]

2. **Farthest Point \(P\)**:
\[ d_{P} = d + r = 5\sqrt{2} + 4 \]

### Step 5: Find the Coordinates of Points \(P\) and \(Q\)
The slope of the line \(CA\) is:
\[ \text{slope} = \frac{2 - (-3)}{7 - 2} = \frac{5}{5} = 1 \]
Thus, the equation of the line through \(C\) is:
\[ y + 3 = 1(x - 2) \implies y = x - 5 \]

Substituting \(y = x - 5\) into the circle equation:
\[ (x - 2)^2 + (x - 5 + 3)^2 = 16 \]
\[ (x - 2)^2 + (x - 2)^2 = 16 \]
\[ 2(x - 2)^2 = 16 \implies (x - 2)^2 = 8 \implies x - 2 = \pm 2\sqrt{2} \]
Thus:
\[ x = 2 \pm 2\sqrt{2} \]

Now substituting back to find \(y\):
1. For \(x = 2 + 2\sqrt{2}\):
\[ y = (2 + 2\sqrt{2}) - 5 = -3 + 2\sqrt{2} \]
So, \(Q = (2 + 2\sqrt{2}, -3 + 2\sqrt{2})\).

2. For \(x = 2 - 2\sqrt{2}\):
\[ y = (2 - 2\sqrt{2}) - 5 = -3 - 2\sqrt{2} \]
So, \(P = (2 - 2\sqrt{2}, -3 - 2\sqrt{2})\).

### Final Answer
- Nearest point \(Q = (2 + 2\sqrt{2}, -3 + 2\sqrt{2})\)
- Farthest point \(P = (2 - 2\sqrt{2}, -3 - 2\sqrt{2})\)