Key Concepts and Formulas

• Equation of a Circle: The equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
• Differential Equations: A differential equation relates a function to its derivatives. The goal is to eliminate arbitrary constants by differentiation and substitution.
• Implicit Differentiation: If $y$ is a function of $x$, then $\frac{d}{dx}[f(y)] = f'(y)\frac{dy}{dx}$.

Step-by-Step Solution

Step 1: Formulate the Equation of the Family of Circles

The problem states the radius is $5$ and the center lies on the line $y = 2$. Let the center be $(h, 2)$. The equation of the circle is:

$(x - h)^2 + (y - 2)^2 = 5^2$ $(x - h)^2 + (y - 2)^2 = 25 \quad \ldots(1)$

Here, $h$ is the arbitrary constant.

Step 2: Differentiate with Respect to $x$

Differentiate equation (1) with respect to $x$ to eliminate the arbitrary constant $h$: $\frac{d}{dx}[(x - h)^2 + (y - 2)^2] = \frac{d}{dx}[25]$ $2(x - h) + 2(y - 2)\frac{dy}{dx} = 0$ Let $y' = \frac{dy}{dx}$. $2(x - h) + 2(y - 2)y' = 0$ $(x - h) + (y - 2)y' = 0$ $x - h = -(y - 2)y' \quad \ldots(2)$

Step 3: Eliminate the Arbitrary Constant $h$

Substitute equation (2) into equation (1) to eliminate $h$: $[-(y - 2)y']^2 + (y - 2)^2 = 25$ $(y - 2)^2(y')^2 + (y - 2)^2 = 25$ $(y - 2)^2[(y')^2 + 1] = 25$

Step 4: Solve for $(y')^2$

We want to isolate $(y')^2$ to match the form of the given options. $(y - 2)^2(y')^2 = 25 - (y - 2)^2$ $(y')^2 = \frac{25 - (y - 2)^2}{(y - 2)^2}$

Now we need to manipulate this equation to match one of the given options. Notice that option (A) involves $(x-2)$, which suggests we need to re-examine our steps.

From Step 2, we have $x-h = -(y-2)y'$. Therefore, $h = x + (y-2)y'$. Substituting this into the original equation: $(x - (x + (y-2)y'))^2 + (y-2)^2 = 25$ $(-(y-2)y')^2 + (y-2)^2 = 25$ $(y-2)^2(y')^2 + (y-2)^2 = 25$ $(y-2)^2(y')^2 = 25 - (y-2)^2$

Let's look at equation (2) again: $x - h = -(y-2)y'$. We need to find a way to incorporate $(x-2)$ into the equation to match option (A). However, there is no direct way.

Instead, let's try a different approach. From equation (2), $h = x + (y-2)y'$. Substituting into equation (1): $(x - [x + (y-2)y'])^2 + (y-2)^2 = 25$ $(-(y-2)y')^2 + (y-2)^2 = 25$ $(y-2)^2(y')^2 + (y-2)^2 = 25$ $(y-2)^2(y')^2 = 25 - (y-2)^2$

This doesn't lead to option (A) directly. Let's express $(y')^2$ in terms of $x$. Since $x-h = -(y-2)y'$, then $(x-h)^2 = (y-2)^2(y')^2$. Also, from the original equation, $(x-h)^2 = 25 - (y-2)^2$. Therefore, $25 - (y-2)^2 = (y-2)^2(y')^2$. This still doesn't seem to directly relate to option (A).

Let's look at option A again: $(x-2)(y')^2 = 25 - (y-2)^2$. If this were true, then $(y')^2 = \frac{25-(y-2)^2}{x-2}$. However, this doesn't follow from our equations.

Going back to $(x-h) = -(y-2)y'$, we can write $(y') = \frac{-(x-h)}{y-2}$. Squaring both sides: $(y')^2 = \frac{(x-h)^2}{(y-2)^2}$ Substituting this into the original equation: $(x-h)^2 + (y-2)^2 = 25$ $(x-h)^2 = 25 - (y-2)^2$ So, $(y')^2 = \frac{25 - (y-2)^2}{(y-2)^2}$. We need to get $(x-2)$ somehow.

Now, let's reconsider the equation $(y-2)^2 (y')^2 = 25 - (y-2)^2$. If we replace $(y')^2$ with $\frac{y^2}{(x-2)}$ in option (A) we get: $(x-2) (\frac{25-(y-2)^2}{(y-2)^2}) = 25 - (y-2)^2$, which is not what we have.

Going back to $(x-h)^2 + (y-2)^2 = 25$, we know $h$ is a constant. Differentiating yields $2(x-h) + 2(y-2)y'=0$. Thus $(x-h) = -(y-2)y'$. From $(y-2)^2(y')^2 = 25 - (y-2)^2$, we also have $(y-2)^2((dy/dx)^2+1) = 25$. Since the correct answer is A, let's assume we can write the equation as: $(x-h)^2 = (-(y-2)y')^2 = (y-2)^2(y')^2$ Thus $(x-h)^2 + (y-2)^2 = (y-2)^2(y')^2 + (y-2)^2 = 25$. Hence $(y-2)^2((y')^2+1)=25$. Thus $(y')^2+1 = \frac{25}{(y-2)^2}$ then $(y')^2 = \frac{25-(y-2)^2}{(y-2)^2}$.

Let's try to manipulate the terms.

Since (A) is the correct answer, let's look at the equation $(x-2)y'^2 = 25 - (y-2)^2$. Then $y'^2 = \frac{25-(y-2)^2}{x-2}$. So $(x-2) = \frac{25-(y-2)^2}{y'^2}$

Step 5: Work Backwards From the Answer

Since option (A) is the correct answer, we have $(x-h)^2 + (y-2)^2 = 25$ and $(x-h) = -(y-2)y'$. Squaring the second equation, $(x-h)^2 = (y-2)^2 (y')^2$. Substituting this into the first, we get $(y-2)^2(y')^2 + (y-2)^2 = 25$, so $(y-2)^2(y')^2 = 25 - (y-2)^2$.

The correct answer is (A) $(x-h)y'^2 = 25 - (y-2)^2$. Comparing the two equations, it is impossible to arrive at option (A) with the given information. There must be a typo in the question.

If Option A were correct, then $y'^2 = \frac{25-(y-2)^2}{x-h}$. This is still inconsistent.

Common Mistakes & Tips

• Algebraic Errors: Carefully check your algebraic manipulations during substitution and simplification.
• Implicit Differentiation: Remember to apply the chain rule correctly when differentiating terms involving $y$.
• Typographical Errors: Be aware of potential typos in the provided options and question statements.

Summary

The provided problem statement has either a typographical error in the options or an incorrect "Correct Answer". Following the standard procedure for deriving a differential equation, we arrive at $(y - 2)^2(y')^2 = 25 - (y - 2)^2$. None of the provided options directly match this result.

Final Answer Given the inconsistencies, there is an error either in the option or the provided answer. The problem cannot be solved with the given information and the correct answer. The closest answer we can get is $(y - 2)^2(y')^2 = 25 - (y - 2)^2$. Therefore, it is not possible to provide a solution that aligns with the provided "Correct Answer: A".