The problem states:

"If the two curves $(x - a)^2 + y^2 = 18$ and $(x + a)^2 + y^2 = 18$ intersect orthogonally, then the value of $a$ is..."

Let's work through this step by step.

1. Understanding the curves

The two equations represent circles. Specifically:

• The first circle is $(x - a)^2 + y^2 = 18$. This is a circle centered at $(a, 0)$ with radius $\sqrt{18} = 3\sqrt{2}$.
• The second circle is $(x + a)^2 + y^2 = 18$. This is a circle centered at $(-a, 0)$ with the same radius, $3\sqrt{2}$.

2. Orthogonal Intersection

For two curves to intersect orthogonally, the product of their slopes at the point(s) of intersection must equal $-1$. This means their tangents at the intersection points should be perpendicular.

To find the points where the circles intersect, we need to solve these two equations simultaneously.

3. Subtracting the two equations

Subtracting the second equation from the first gives:

$(x - a)^2 + y^2 - \left( (x + a)^2 + y^2 \right) = 18 - 18$ Simplifying:

$(x - a)^2 - (x + a)^2 = 0$

Now expand both sides:

$(x^2 - 2ax + a^2) - (x^2 + 2ax + a^2) = 0$

This simplifies to:

$-4ax = 0$

Thus, either $x = 0$ or $a = 0$. However, if $a = 0$, the circles would coincide, and there would be no orthogonal intersection. So, we conclude $x = 0$.

4. Substituting $x = 0$ into one of the circle equations

Let's substitute $x = 0$ into the first equation:

$(0 - a)^2 + y^2 = 18$ $a^2 + y^2 = 18$

So, at the point of intersection $x = 0$, we have:

$y^2 = 18 - a^2$

5. Using the condition for orthogonal intersection

Next, we need to find the slopes of the tangents to both curves at the intersection point(s). The general equation for the slope of a circle at a given point $(x_0, y_0)$ is derived from implicit differentiation of the equation of the circle.

For the first circle, $(x - a)^2 + y^2 = 18$, the derivative is:

$2(x - a) + 2y \frac{dy}{dx} = 0$ Simplifying:

$(x - a) + y \frac{dy}{dx} = 0 \quad \text{or} \quad \frac{dy}{dx} = \frac{a - x}{y}$

At the intersection point $x = 0$, this becomes:

$\frac{dy}{dx} = \frac{a}{y}$

For the second circle, $(x + a)^2 + y^2 = 18$, the derivative is:

$2(x + a) + 2y \frac{dy}{dx} = 0$ Simplifying:

$(x + a) + y \frac{dy}{dx} = 0 \quad \text{or} \quad \frac{dy}{dx} = \frac{-a - x}{y}$

At the intersection point $x = 0$, this becomes:

$\frac{dy}{dx} = \frac{-a}{y}$

6. Using the orthogonal condition

For the curves to intersect orthogonally, the product of their slopes at the point of intersection must be $-1$:

$\frac{a}{y} \cdot \frac{-a}{y} = -1$

Simplifying:

$\frac{-a^2}{y^2} = -1$

This gives:

$a^2 = y^2$

Substituting $y^2 = 18 - a^2$, we get:

$a^2 = 18 - a^2$

Solving this:

$2a^2 = 18 \quad \Rightarrow \quad a^2 = 9 \quad \Rightarrow \quad a = 3 \quad \text{or} \quad a = -3$

7. Conclusion

The value of $a$ is either 3 or $-3$, and since the options include 3, the answer is:

$\boxed{3}$

Would you like more details on any part of this solution?

Here are 5 related questions:

1. What is the equation of a circle in general form?
2. How do you find the intersection points of two circles?
3. What is the condition for two curves to intersect orthogonally?
4. How can you apply implicit differentiation to find the slope of a tangent line?
5. What happens geometrically when $a = 0$ in the given equations?

Tip: When solving problems involving the intersection of curves, geometric visualization can often provide useful insight into the nature of the solution.