Question

The angle between the circles \(x^2+y^2−4x−6y−3=0\), \(x^2+y^2+8x−4y+11=0\) is \(\frac{\pi}{2}\), then the value of K is?

• A. 11
• B. 10
• C. -15
• D. 14

Answer 1

The Correct Option is C

To find the value of \(k\) for which the circles \(C_1: x^2 + y^2 - 4x - 6y + k = 0\) and \(C_2: x^2 + y^2 + 8x - 4y + 11 = 0\) are orthogonal, we proceed as follows:

1. Determining the Centers of the Circles:
For circle \(C_1: x^2 + y^2 - 4x - 6y + k = 0\), rewrite in standard form by completing the square:

For \(x\): \(x^2 - 4x = (x - 2)^2 - 4\)
For \(y\): \(y^2 - 6y = (y - 3)^2 - 9\)
So:

\( (x - 2)^2 - 4 + (y - 3)^2 - 9 + k = 0 \)
\( (x - 2)^2 + (y - 3)^2 = 13 - k \)
The center is \(O_1(2, 3)\).

For circle \(C_2: x^2 + y^2 + 8x - 4y + 11 = 0\):

For \(x\): \(x^2 + 8x = (x + 4)^2 - 16\)
For \(y\): \(y^2 - 4y = (y - 2)^2 - 4\)
So:

\( (x + 4)^2 - 16 + (y - 2)^2 - 4 + 11 = 0 \)
\( (x + 4)^2 + (y - 2)^2 = 9 \)
The center is \(O_2(-4, 2)\).

2. Calculating the Radii:
For \(C_1\), the radius is:

\( r_1 = \sqrt{13 - k} \)
For \(C_2\), the radius is:

\( r_2 = \sqrt{9} = 3 \)

3. Orthogonality Condition:
Two circles are orthogonal if the angle between them is \(\pi/2\). For circles in the form \(x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0\) and \(x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0\), the orthogonality condition is:

\( 2 g_1 g_2 + 2 f_1 f_2 = c_1 + c_2 \)
Identify coefficients:

For \(C_1\): \(2g_1 = -4 \implies g_1 = -2\), \(2f_1 = -6 \implies f_1 = -3\), \(c_1 = k\)
For \(C_2\): \(2g_2 = 8 \implies g_2 = 4\), \(2f_2 = -4 \implies f_2 = -2\), \(c_2 = 11\)

4. Applying the Orthogonality Condition:
Substitute into the orthogonality condition:

\( 2 (-2) (4) + 2 (-3) (-2) = k + 11 \)
\( -16 + 12 = k + 11 \)
\( -4 = k + 11 \)
\( k = -15 \)

5. Verifying the Radius:
With \(k = -15\), check the radius of \(C_1\):

\( r_1 = \sqrt{13 - (-15)} = \sqrt{13 + 15} = \sqrt{28} = 2\sqrt{7} \)
This is a positive real number, confirming the circle exists.

Final Answer:
The value of \(k\) for which the circles are orthogonal is \(-15\).