Step 1: Understanding the Concept:
Two circles intersect orthogonally if the angle between their tangents at the points of intersection is \(90^\circ\).
This geometric condition translates to a specific algebraic relationship involving the coefficients of their general equations.
Step 2: Key Formula or Approach:
The general equation of a circle is written as \(x^2 + y^2 + 2gx + 2fy + c = 0\).
For two circles with equations \(x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0\) and \(x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0\), the condition for orthogonal intersection is:
\[ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \]
Step 3: Detailed Explanation:
First, we must express the given equations in the standard general form where the coefficients of \(x^2\) and \(y^2\) are exactly \(1\).
First Circle: The given equation is \(2x^2 + 2y^2 - 4x + 6y = 3\).
We divide the entire equation by \(2\) to normalize it:
\[ x^2 + y^2 - 2x + 3y - \frac{3}{2} = 0 \]
Comparing this with the general form \(x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0\), we extract the constants:
\(2g_1 = -2 \Rightarrow g_1 = -1\)
\(2f_1 = 3 \Rightarrow f_1 = \frac{3}{2}\)
\(c_1 = -\frac{3}{2}\)
Second Circle: The given equation is already in the general form: \(x^2 + y^2 + kx + y = 0\).
Comparing this with \(x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0\), we get:
\(2g_2 = k \Rightarrow g_2 = \frac{k}{2}\)
\(2f_2 = 1 \Rightarrow f_2 = \frac{1}{2}\)
\(c_2 = 0\)
Now, we apply the condition for orthogonality:
\[ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \]
Substitute the values we have identified:
\[ 2(-1)\left(\frac{k}{2}\right) + 2\left(\frac{3}{2}\right)\left(\frac{1}{2}\right) = -\frac{3}{2} + 0 \]
Simplifying the terms:
\[ -k + \frac{3}{2} = -\frac{3}{2} \]
To solve for \(k\), isolate the variable:
\[ -k = -\frac{3}{2} - \frac{3}{2} \]
\[ -k = -\frac{6}{2} \]
\[ -k = -3 \]
Therefore, multiplying by \(-1\) yields \(k = 3\).
Step 4: Final Answer:
The value of \(k\) is \(3\).