Here is a more elaborate, clear, and educational solution to the problem.

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1. Understanding the Problem and Key Strategy

We are given the general equation of a circle $x^2 + y^2 + Ax + By + C = 0$. Our goal is to find the value of $A+C$. We are provided with two crucial pieces of information about this circle:

1. It passes through the point $(0, 6)$.
2. It touches the parabola $y = x^2$ at the point $(2, 4)$.

The second condition, "touching the parabola $y = x^2$ at $(2, 4)$," is the cornerstone of our solution strategy. When two curves touch at a point, they share a common tangent line at that point. This geometric insight allows us to employ a powerful concept from coordinate geometry: the Family of Circles Touching a Line at a Point. This method significantly simplifies the problem compared to trying to solve a system of equations by substituting points directly into the general circle equation, especially when tangency is involved.

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2. Key Concept: Family of Circles Touching a Line at a Point

Let $L = 0$ be the equation of a line, and let $(x_1, y_1)$ be a point that lies on this line. The equation of any circle that touches the line $L=0$ at the point $(x_1, y_1)$ can be represented by the general form: $(x - x_1)^2 + (y - y_1)^2 + \lambda L = 0$ where $\lambda$ (lambda) is a real parameter. Each unique value of $\lambda$ corresponds to a unique circle in this family.

Why does this formula work?

• The term $(x - x_1)^2 + (y - y_1)^2 = 0$ represents a "point circle" (a degenerate circle with zero radius) centered at $(x_1, y_1)$. This essentially defines the point of tangency.
• The term $L=0$ represents the tangent line itself.
• The general equation $S + \lambda L = 0$ (where $S=0$ is a circle and $L=0$ is a line) represents a family of circles passing through the intersection points of the circle $S=0$ and the line $L=0$. In our specific case, the "circle" $S=0$ is the point circle $(x-x_1)^2 + (y-y_1)^2=0$. Since this point $(x_1, y_1)$ lies on the line $L=0$, the intersection of the point circle and the line is precisely this single point $(x_1, y_1)$. This means that any curve represented by this family equation will touch the line $L=0$ at the point $(x_1, y_1)$, sharing the tangent $L=0$.

Our strategy will be to:

1. Find the equation of the common tangent line $L=0$ at the point $(2, 4)$.
2. Form the general equation of the family of circles touching $L=0$ at $(2, 4)$.
3. Use the additional condition that the circle passes through $(0, 6)$ to find the specific value of $\lambda$.
4. Substitute $\lambda$ back into the circle's equation and identify the coefficients $A$ and $C$.

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3. Step 1: Finding the Equation of the Common Tangent Line

Since the circle touches the parabola $y = x^2$ at the point $(2, 4)$, both curves share the same tangent line at this point. Our first task is to find the equation of this common tangent line.

1. Calculate the slope of the tangent to the parabola: The equation of the parabola is $y = x^2$. To find the slope of the tangent at any point, we use differentiation: $\frac{dy}{dx} = \frac{d}{dx}(x^2) = 2x$

2. Determine the slope at the specific point of tangency $(2, 4)$: We substitute the $x$-coordinate of the tangency point, $x=2$, into the derivative: $m = \left.\frac{dy}{dx}\right|_{(2,4)} = 2(2) = 4$ So, the slope of the common tangent line at $(2, 4)$ is $m = 4$.

3. Write the equation of the tangent line: We use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$, with the point $(x_1, y_1) = (2, 4)$ and the slope $m=4$: $y - 4 = 4(x - 2)$ $y - 4 = 4x - 8$ To match the general form $L = Ax + By + C = 0$, we rearrange the equation: $4x - y - 4 = 0$ This is the equation of the common tangent line, which we will denote as $L=0$.

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4. Step 2: Forming the Equation of the Family of Circles

Now we apply the key concept introduced earlier. The circle we are looking for touches the line $L = 4x - y - 4 = 0$ at the point $(x_1, y_1) = (2, 4)$.

Using the formula for the family of circles: $(x - x_1)^2 + (y - y_1)^2 + \lambda L = 0$: $(x - 2)^2 + (y - 4)^2 + \lambda(4x - y - 4) = 0$ This equation represents every possible circle that touches the line $4x - y - 4 = 0$ at the point $(2, 4)$. Our specific circle is one of these, and its identity is determined by a unique value of $\lambda$.

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5. Step 3: Determining the Specific Circle using the Given Point

The problem states that our specific circle also passes through the point $(0, 6)$. We can use this information to find the unique value of $\lambda$ that corresponds to our circle.

1. Substitute the coordinates $(0, 6)$ into the family of circles equation: Since the circle passes through $(0, 6)$, these coordinates must satisfy its equation: $(0 - 2)^2 + (6 - 4)^2 + \lambda(4(0) - 6 - 4) = 0$

2. Simplify and solve for $\lambda$: $(-2)^2 + (2)^2 + \lambda(0 - 6 - 4) = 0$ $4 + 4 + \lambda(-10) = 0$ $8 - 10\lambda = 0$ $10\lambda = 8$ $\lambda = \frac{8}{10} = \frac{4}{5}$ Thus, the parameter $\lambda = \frac{4}{5}$ uniquely defines the circle that satisfies all given conditions.

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6. Step 4: Identifying Coefficients A and C

Now that we have the value of $\lambda$, we substitute it back into the family of circles equation and expand it. Our goal is to match this expanded equation with the general form $x^2 + y^2 + Ax + By + C = 0$ to identify the coefficients $A$ and $C$.

1. Substitute $\lambda = \frac{4}{5}$ into the equation of the circle: $(x - 2)^2 + (y - 4)^2 + \frac{4}{5}(4x - y - 4) = 0$

2. Expand the squared binomial terms: Recall that $(a-b)^2 = a^2 - 2ab + b^2$. $(x^2 - 4x + 4) + (y^2 - 8y + 16) + \frac{4}{5}(4x - y - 4) = 0$

3. Combine constant terms and distribute $\frac{4}{5}$: $x^2 + y^2 - 4x - 8y + (4 + 16) + \frac{16}{5}x - \frac{4}{5}y - \frac{16}{5} = 0$ $x^2 + y^2 - 4x - 8y + 20 + \frac{16}{5}x - \frac{4}{5}y - \frac{16}{5} = 0$

4. Group terms by powers of $x$, $y$, and constants to match the general form: $x^2 + y^2 + \left(-4 + \frac{16}{5}\right)x + \left(-8 - \frac{4}{5}\right)y + \left(20 - \frac{16}{5}\right) = 0$

5. Calculate the coefficients $A$, $B$, and $C$:

   • For the $x$ term (coefficient $A$): $A = -4 + \frac{16}{5} = -\frac{20}{5} + \frac{16}{5} = -\frac{4}{5}$
   • For the $y$ term (coefficient $B$): $B = -8 - \frac{4}{5} = -\frac{40}{5} - \frac{4}{5} = -\frac{44}{5}$
   • For the constant term (coefficient $C$): $C = 20 - \frac{16}{5} = \frac{100}{5} - \frac{16}{5} = \frac{84}{5}$

JEE Tip: Always double-check your arithmetic, especially when dealing with fractions. A small error here can lead to an incorrect final answer.

The problem asks for the value of $A+C$. $A+C = \left(-\frac{4}{5}\right) + \left(\frac{84}{5}\right)$ $A+C = \frac{-4 + 84}{5} = \frac{80}{5}$ $A+C = 16$

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7. Final Answer and Conclusion

The value of $A+C$ is $16$.

The final answer is $\boxed{\text{16}}$.

This problem beautifully demonstrates the power and elegance of using the "Family of Circles" concept in coordinate geometry. By recognizing that tangency implies a common tangent line, we could construct the general equation of the circle very efficiently. This approach avoids complex algebraic manipulations that might arise from other methods (like finding the center and radius directly). This technique is a crucial tool for solving problems involving circles and tangency in competitive exams like JEE.