Without change of axes the origin is shifted to (h, k), then from the equation `x^(2)+y^(2)-4x+6y-7=0`, the term containing linear powers are missing, then point (h, k) is

Options

A. (a) (3, 2)
B. (b) (-3, 2)
C. (c) (2, -3)
D. (d) (-2, -3)

Answer

To solve the problem, we need to shift the origin to the point (h, k) and analyze the given equation \(x^2 + y^2 - 4x + 6y - 7 = 0\) to find the values of h and k such that the linear terms disappear.

### Step-by-step Solution:

1. **Understand the Transformation**:
We are shifting the origin to the point (h, k). This means we will substitute \(x\) and \(y\) in the equation with \(x' + h\) and \(y' + k\) respectively, where \(x'\) and \(y'\) are the new coordinates after the shift.

\[ x = x' + h \quad \text{and} \quad y = y' + k \]

2. **Substituting in the Equation**:
Substitute \(x\) and \(y\) into the equation:

\[ (x' + h)^2 + (y' + k)^2 - 4(x' + h) + 6(y' + k) - 7 = 0 \]

3. **Expanding the Equation**:
Expand the equation:

\[ (x'^2 + 2hx' + h^2) + (y'^2 + 2ky' + k^2) - 4x' - 4h + 6y' + 6k - 7 = 0 \]

Combine like terms:

\[ x'^2 + y'^2 + (2h - 4)x' + (2k + 6)y' + (h^2 + k^2 - 4h + 6k - 7) = 0 \]

4. **Identifying Linear Terms**:
For the linear terms in \(x'\) and \(y'\) to be missing, the coefficients of \(x'\) and \(y'\) must equal zero:

\[ 2h - 4 = 0 \quad \text{and} \quad 2k + 6 = 0 \]

5. **Solving for h and k**:
Solve the equations:

- From \(2h - 4 = 0\):
\[ 2h = 4 \implies h = 2 \]

- From \(2k + 6 = 0\):
\[ 2k = -6 \implies k = -3 \]

6. **Conclusion**:
The point (h, k) is:

\[ (h, k) = (2, -3) \]

### Final Answer:
The point (h, k) is \( (2, -3) \).

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