If the points of intersection of the ellipses x² + 2y² - 6x - 12y + 23 = 0 and 4x² + 2y² - 20x - 12y

Q: Why do we use the family of curves $S_1 + \lambda S_2 = 0$? Can't we just find the intersection points directly?

That is a fantastic question! If you try to solve the system of two ellipses directly, you will end up with a quartic equation that is algebraically nightmarish. The 'Family of Curves' method is a beautiful shortcut. By using the parameter $\lambda$, we are essentially creating a new curve that is guaranteed to pass through the intersection points of the original two, without ever needing to know the coordinates of those points. It is the elegant way, not the brute-force way.

Q: Why is the coefficient of $xy$ zero in our final equation?

Great observation! In the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$, there is no $xy$ term. In our specific problem, both given ellipses lack an $xy$ term. When we combine them using $S_1 + \lambda S_2 = 0$, the resulting equation also lacks an $xy$ term. If there were an $xy$ term, the curve would be a rotated conic, not a circle. We are lucky that the geometry here is aligned with the axes!

Q: Why do we divide the entire equation by 3 at the end?

To find the center $(a, b)$ and the radius $r$ of a circle, we must compare our equation to the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$. In this standard form, the coefficients of $x^2$ and $y^2$ must be exactly 1. Our equation had a coefficient of 3 for both, so dividing by 3 is the necessary step to normalize the equation and extract the correct values for $g$, $f$, and $c$.

Question

If the points of intersection of the ellipses $x^2 + 2y^2 - 6x - 12y + 23 = 0$ and $4x^2 + 2y^2 - 20x - 12y + 35 = 0$ lie on a circle of radius $r$ and centre $(a, b)$, then the value of $ab + 18r^2$ is

Options:

(A) 52

(B) 53

(C) 51

(D) 55

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a coordinate plane, watching two ellipses, $S_1$ and $S_2$, dancing toward each other. They are distinct, with different centers and different eccentricities, yet they collide at four specific points.

The problem asks us to find the circle that passes through these four points. At first glance, you might be tempted to solve for the intersection points directly. I urge you: resist that temptation!

That path leads to a swamp of quartic equations and algebraic despair. Instead, we will use the **Family of Curves**—a concept that is as powerful as it is elegant.

### The Master Key: The Family of Curves

When we write the equation $S_1 + \lambda S_2 = 0$, we are not just writing a random equation. We are invoking a geometric truth.

Any curve that passes through the intersection of $S_1$ and $S_2$ must satisfy this linear combination. By introducing the parameter $\lambda$, we are essentially creating a 'slider' that allows us to morph through every possible curve passing through those four points.

Our goal is to find the specific value of $\lambda$ that forces this curve to be a circle.

### Forcing the Circle

A general second-degree equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ is only a circle if $A = C$ and $B = 0$.

Let us look at our combined equation:

$$ (1 + 4\lambda)x^2 + (2 + 2\lambda)y^2 - (6 + 20\lambda)x - (12 + 12\lambda)y + (23 + 35\lambda) = 0 $$

For this to be a circle, the coefficient of $x^2$ must equal the coefficient of $y^2$. This gives us the simple, beautiful condition:

$$ 1 + 4\lambda = 2 + 2\lambda $$

Solving this, we find $2\lambda = 1$, or $\lambda = \frac{1}{2}$. This is the moment of clarity. We have found the exact parameter that transforms our general conic into the specific circle we seek.

### The Final Assembly

With $\lambda = \frac{1}{2}$ in hand, we substitute it back into our equation. The math unfolds cleanly:

$$ 3x^2 + 3y^2 - 16x - 18y + \frac{81}{2} = 0 $$

Now, we normalize. We divide by 3 to get:

$$ x^2 + y^2 - \frac{16}{3}x - 6y + \frac{27}{2} = 0 $$

Comparing this to the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$, we identify $g = -\frac{8}{3}$ and $f = -3$. The center $(a, b)$ is $(-g, -f)$, which is $(\frac{8}{3}, 3)$.

Finally, we calculate the radius squared:

$$ r^2 = g^2 + f^2 - c = \left(-\frac{8}{3}\right)^2 + (-3)^2 - \frac{27}{2} = \frac{64}{9} + 9 - \frac{27}{2} = \frac{47}{18} $$

The final calculation, $ab + 18r^2$, becomes:

$$ \left(\frac{8}{3}\right)(3) + 18\left(\frac{47}{18}\right) = 8 + 47 = \mathbf{55} $$

We have arrived at the destination. It is a testament to the beauty of coordinate geometry that such a complex intersection can be resolved with such precision. Keep this method in your toolkit; it is the mark of a true problem solver.

Animated Solution for Mathematics - Circles: If the points of intersection of the ellipses $x^{2} + 2 y^{2} - 6 x - 12 y + 23 = 0$ and $4 x^{2} + 2 y^{2} - 20 x - 12 y + 35 = 0$ lie on a circle of radius $r$ and centre $(a, b)$ , then the value of $ab + 18 r^{2}$ is

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Visualized Solution

Visualizing the Ellipses $S_{1}$ and $S_{2}$

The Points of Intersection

Family of Curves

Setting up the Combined Equation

Grouping the Terms

Condition for a Circle

Solving for $λ$

Substituting $λ$

Standardizing the Circle Equation

Extracting Center $(a, b)$

Calculating Radius Squared $r^{2}$

Final Calculation: $ab + 18 r^{2}$

The Sigma Insight: Family of Circles

Analyzing the Setup

The Master Key

Forcing the Circle

The Final Assembly