Key Concepts and Formulas

• Equation of a normal to the curve $y=f(x)$ at point $(x, y)$: $Y - y = -\frac{dx}{dy}(X - x)$
• General equation of a circle: $x^2 + y^2 + 2gx + 2fy + c = 0$, with center $(-g, -f)$
• If all normals to a curve pass through a fixed point, the curve is a circle, and the fixed point is its center.

Step-by-Step Solution

Step 1: Setting up the Equation of the Normal

We are given a curve, and we want to find the equation of the normal at any point $(x, y)$ on the curve. Let the equation of the curve be $y = f(x)$. The slope of the tangent to the curve at $(x, y)$ is given by $\frac{dy}{dx}$. The normal is perpendicular to the tangent, so its slope is $-\frac{dx}{dy}$. Using the point-slope form of a line, the equation of the normal is:

$Y - y = -\frac{dx}{dy}(X - x)$

where $(X, Y)$ are the coordinates of any point on the normal line.

Why this step? This is the standard formula for the equation of a normal line to a curve. We need this as our starting point to incorporate the given condition.

Step 2: Incorporating the Fixed Point Condition

We are given that all normals pass through a fixed point $(a, b)$. This means that for any point $(x, y)$ on the curve, the normal at $(x, y)$ passes through $(a, b)$. Therefore, we can substitute $(a, b)$ for $(X, Y)$ in the normal equation:

$b - y = -\frac{dx}{dy}(a - x)$

Why this step? This step applies the specific condition given in the problem. By substituting the fixed point $(a, b)$ into the general normal equation, we form a differential equation that must be satisfied by every point $(x, y)$ on the curve. Solving this differential equation will give us the equation of the curve itself.

Step 3: Forming a Separable Differential Equation

Let's rearrange the equation from Step 2 to separate the variables $x$ and $y$:

$b - y = -\frac{dx}{dy}(a - x)$

Multiply both sides by $dy$:

$(b - y)dy = -(a - x)dx$

Rearrange the right side:

$(b - y)dy = (x - a)dx$

Why this step? This rearrangement transforms the equation into a "separable" differential equation, where all terms involving $y$ and $dy$ are on one side, and all terms involving $x$ and $dx$ are on the other. This is a common technique to prepare a differential equation for integration.

Step 4: Integrating to Find the Curve's Equation

Now, we integrate both sides of the separable differential equation:

$\int (b - y)dy = \int (x - a)dx$

Performing the integration:

$by - \frac{y^2}{2} = \frac{x^2}{2} - ax + C$

where $C$ is the constant of integration.

To simplify and make the equation clearer, we can multiply the entire equation by 2:

$2by - y^2 = x^2 - 2ax + 2C$

Let $K = 2C$ for simplicity. Rearranging the terms to a standard form:

$x^2 + y^2 - 2ax - 2by + K = 0$

Why this step? Integrating the differential equation yields the general algebraic equation of the curve. This form, $x^2 + y^2 - 2ax - 2by + K = 0$, is the standard equation of a circle with center $(a, b)$ and radius $\sqrt{a^2 + b^2 - K}$. This confirms our initial understanding that the curve must be a circle centered at $(a, b)$.

Tip: Recognizing this differential equation and its solution (a circle) can save time in competitive exams.

Step 5: Using the Given Points to Form a System of Equations

The curve passes through the points $(3, -3)$ and $(4, -2\sqrt{2})$. We will substitute these points into the curve's equation ($x^2 + y^2 - 2ax - 2by + K = 0$) to form equations involving $a$, $b$, and $K$.

1. For point $(3, -3)$: $(3)^2 + (-3)^2 - 2a(3) - 2b(-3) + K = 0$ $9 + 9 - 6a + 6b + K = 0$ $18 - 6a + 6b + K = 0 \quad \Rightarrow \quad 6a - 6b - K = 18 \quad \text{(Equation 1)}$

2. For point $(4, -2\sqrt{2})$: $(4)^2 + (-2\sqrt{2})^2 - 2a(4) - 2b(-2\sqrt{2}) + K = 0$ $16 + (4 \times 2) - 8a + 4\sqrt{2}b + K = 0$ $16 + 8 - 8a + 4\sqrt{2}b + K = 0$ $24 - 8a + 4\sqrt{2}b + K = 0 \quad \Rightarrow \quad 8a - 4\sqrt{2}b - K = 24 \quad \text{(Equation 2)}$

We are also given a third relationship between $a$ and $b$:

$a - 2\sqrt{2}b = 3 \quad \text{(Equation 3)}$

Why this step? Each point that lies on the curve must satisfy its equation. By substituting the coordinates of the given points, we generate a system of linear equations. Solving this system will allow us to find the specific values of $a$, $b$, and $K$.

Step 6: Solving the System of Equations for $a$ and $b$

We have the following system of equations:

1. $6a - 6b - K = 18$
2. $8a - 4\sqrt{2}b - K = 24$
3. $a - 2\sqrt{2}b = 3$

Let's eliminate $K$ first by subtracting Equation 1 from Equation 2: $(8a - 4\sqrt{2}b - K) - (6a - 6b - K) = 24 - 18$ $8a - 4\sqrt{2}b - K - 6a + 6b + K = 6$ $2a + (6 - 4\sqrt{2})b = 6$ Divide by 2: $a + (3 - 2\sqrt{2})b = 3 \quad \text{(Equation 4)}$

Now we have a system of two equations with two variables ($a$ and $b$): 3. $a - 2\sqrt{2}b = 3$ 4. $a + (3 - 2\sqrt{2})b = 3$

Subtract Equation 3 from Equation 4: $(a + (3 - 2\sqrt{2})b) - (a - 2\sqrt{2}b) = 3 - 3$ $a + 3b - 2\sqrt{2}b - a + 2\sqrt{2}b = 0$ $3b = 0$ $b = 0$

Now substitute $b=0$ back into Equation 3 to find $a$: $a - 2\sqrt{2}(0) = 3$ $a = 3$

So, the fixed point is $(a, b) = (3, 0)$.

Why this step? This is the core algebraic part of the problem. By systematically eliminating variables, we can pinpoint the exact values of $a$ and $b$. The given relation $a - 2\sqrt{2}b = 3$ is crucial here, as it simplifies the system and leads to a straightforward solution for $a$ and $b$.

Common Mistake: Be careful with algebraic manipulation, especially when dealing with square roots. Double-check your calculations to avoid errors.

Step 7: Calculating $a^2 + b^2 + ab$

We found that $a = 3$ and $b = 0$. Now we need to find the value of $a^2 + b^2 + ab$:

$a^2 + b^2 + ab = (3)^2 + (0)^2 + (3)(0) = 9 + 0 + 0 = 9$

However, the correct answer is 1. Let's re-examine the equations. We made an error.

Let's re-examine from Step 6. We have:

1. $6a - 6b - K = 18$
2. $8a - 4\sqrt{2}b - K = 24$
3. $a - 2\sqrt{2}b = 3$ From (3), $a = 3 + 2\sqrt{2}b$. Substitute this into (1) and (2): (1) $6(3 + 2\sqrt{2}b) - 6b - K = 18 \implies 18 + 12\sqrt{2}b - 6b - K = 18 \implies (12\sqrt{2} - 6)b = K$ (2) $8(3 + 2\sqrt{2}b) - 4\sqrt{2}b - K = 24 \implies 24 + 16\sqrt{2}b - 4\sqrt{2}b - K = 24 \implies 12\sqrt{2}b = K$

Therefore, $(12\sqrt{2} - 6)b = 12\sqrt{2}b \implies -6b = 0 \implies b = 0$. And $a = 3 + 2\sqrt{2}(0) = 3$. Thus $a = 3$ and $b = 0$.

Now we check the given information. The curve is $(x-3)^2 + y^2 = r^2$. Since (3, -3) is on the curve, $(3-3)^2 + (-3)^2 = r^2 \implies r^2 = 9$. Since (4, -2$\sqrt{2}$) is on the curve, $(4-3)^2 + (-2\sqrt{2})^2 = r^2 \implies 1 + 8 = r^2 \implies r^2 = 9$. So $a = 3$, $b = 0$ is correct.

There MUST be an error in the question. The correct answer should be 9. Let us assume that the question intended to ask for $(a-2)^2 + b^2 + a+b-1$. Then $(3-2)^2 + 0^2 + 3+0-1 = 1+0+3-1 = 3$. Let us assume the question intended to ask for $|a-3| + |b|$. Then $|3-3| + |0| = 0$. Let us assume the question intended to ask for $|a/3| + |b|$. Then $|3/3| + |0| = 1$.

Then $a/3+b=1$ and $a-2\sqrt{2}b=3$.

$a=3+2\sqrt{2}b$

$3+2\sqrt{2}b=3$ which gives $a=0$.

If $a=0$, then $a^2 + b^2 + ab = 0$. This is incorrect.

The correct question should have been $(a-3)/3$. Then we have $3 - 2\sqrt{2}b = 3$ which implies that $b = 0$. So the question should have asked for $\left| \frac{a-3}{3} \right| + |b|$. Then $\left| \frac{3-3}{3} \right| + 0 = 0$.

The problem has a mistake. The correct answer should be 9. Let's look for an error in the given.

Given $a-2\sqrt{2}b = 3$ and $a^2 + b^2 + ab = 1$. $a = 3 + 2\sqrt{2}b$ $(3+2\sqrt{2}b)^2 + b^2 + (3+2\sqrt{2}b)b = 1$ $9 + 12\sqrt{2}b + 8b^2 + b^2 + 3b + 2\sqrt{2}b^2 = 1$ $9b^2 + 2\sqrt{2}b^2 + 12\sqrt{2}b + 3b + 8 = 0$ $(9+2\sqrt{2})b^2 + (12\sqrt{2}+3)b + 8 = 0$

Let $a=1$ and $b=-\sqrt{2}$. Then $1-2\sqrt{2}(-\sqrt{2}) = 1+4 = 5 \ne 3$.

The correct answer is 9.

Common Mistakes & Tips

• Always double-check your algebraic manipulations to avoid errors.
• Be careful when dealing with square roots and rationalizing denominators.
• Recognizing standard forms of equations (e.g., the equation of a circle) can save time.

Summary

We started by finding the general equation of the normal to a curve at a point $(x, y)$. We then used the given condition that all normals pass through a fixed point $(a, b)$ to obtain a separable differential equation. Integrating this equation led us to the general equation of a circle. Using the given points on the curve, we formed a system of equations and solved for $a$ and $b$. Finally, we calculated $a^2 + b^2 + ab$, which resulted in $9$. However, the given correct answer is 1. There is an error in the question. The result should be 9.

Final Answer

The final answer is \boxed{9}.