Question

The value of \(k\), if the circles \(2x^{2} + 2y^{2} - 4x + 6y = 3\) and \(x^{2} + y^{2} + kx + y = 0\) cut orthogonally is

• A. 2
• B. 3
• C. 4
• D. 5
• E. 1

Hint:

Always ensure the coefficients of \(x^2\) and \(y^2\) are equal to 1 before identifying \(g\), \(f\), and \(c\). Failing to divide by 2 in the first equation is a common mistake.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Two circles cut each other orthogonally if the tangents at their point of intersection are perpendicular. This condition is expressed in terms of their general equation coefficients.

Step 2: Key Formula or Approach:
For two circles \(x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0\) and \(x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0\), the condition for orthogonality is: \[ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \]

Step 3: Detailed Explanation:
1. Normalize the first circle: \(x^2 + y^2 - 2x + 3y - 1.5 = 0\). Here, \(g_1 = -1\), \(f_1 = 1.5\), \(c_1 = -1.5\).
2. Identify coefficients of the second circle: \(g_2 = k/2\), \(f_2 = 1/2\), \(c_2 = 0\).
3. Apply the orthogonality condition: \[ 2(-1)(\frac{k}{2}) + 2(1.5)(\frac{1}{2}) = -1.5 + 0 \] \[ -k + 1.5 = -1.5 \]
4. Solve for \(k\): \(-k = -3 \implies k = 3\). Wait, let's re-verify: \(2g_1g_2 + 2f_1f_2 = c_1 + c_2 \implies -k + 1.5 = -1.5 \implies k = 3\). Re-checking normalization: \(2x^2 + 2y^2 - 4x + 6y - 3 = 0 \to x^2 + y^2 - 2x + 3y - 3/2 = 0\). Correct.

Step 4: Final Answer
The value of \(k\) is 3.