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 $-15$.