Q
Analytical
Intersection
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 :
A) $53$ B) $52$ C) $55$ D) $51$
1. Family of Curves (S + λL = 0)
When two curves $S_1 = 0$ and $S_2 = 0$ intersect, any curve passing through their intersection points can be expressed as $S_1 + \lambda(S_2 – S_1) = 0$ or $S_1 + \lambda S_2 = 0$. In this problem, $E_2 – E_1$ represents a straight line (the common chord) because the $y^2$ terms cancel out. This reduces the problem from finding intersection points to finding a specific $\lambda$ that satisfies the circle condition.
2. Circle Conditions in 2D Geometry
A general second-degree equation $Ax^2 + By^2 + 2Hxy + 2Gx + 2Fy + C = 0$ represents a circle if and only if $A = B$ (and $A \neq 0$) and $H = 0$. By forcing the coefficients of $x^2$ and $y^2$ to be equal in the family of curves, we ensure the resulting curve is a circle. This is a powerful technique to avoid calculating messy intersection coordinates.
3. Properties of Ellipses
An ellipse is defined as the locus of points whose distances from two fixed points (foci) sum to a constant. Algebraically, it is $x^2/a^2 + y^2/b^2 = 1$. In this problem, the ellipses are shifted from the origin. Finding the centre of a circle passing through four points of intersection of two conics is a classic problem in coordinate geometry.
4. Radius and Centre Calculation
For a circle $x^2 + y^2 + 2gx + 2fy + c = 0$, the centre is $(-g, -f)$ and the radius $r$ is given by $\sqrt{g^2 + f^2 – c}$. Accuracy in this step is vital because the final answer requires $r^2$. Errors in the constant term $c$ or the coefficients of linear terms will directly affect the radius squared.
1
Can two ellipses intersect at more than 4 points?
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No, according to Bézout’s theorem, two general second-degree curves can intersect at a maximum of $2 \times 2 = 4$ points.
2
Why did we assume the centre is $(a, b)$?
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The question explicitly states that the circle has centre $(a, b)$ and radius $r$. We solve the equation to find these specific values.
3
What if $A \neq B$ in the final equation?
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If $A \neq B$, the curve is not a circle; it could be an ellipse, parabola, or hyperbola. We chose $\lambda$ specifically to make $A = B$.
4
Is $r^2$ always positive?
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For a real circle, $r^2 = g^2 + f^2 – c$ must be greater than 0. If it were zero, it would be a point circle.
5
Can I use the substitution method for x and y?
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You can, but it’s much harder. Solving for $x$ gives two values, and substituting them into the ellipse gives four $y$ values. Solving for a circle through 4 points is algebraically heavy.
6
What is the common chord in this problem?
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The common chord is the line $3x^2 – 14x + 12 = 0$. Wait, that’s not a line. The common chord exists only if the $x^2$ terms also match. Here, they intersect at vertical lines $x=constant$.
7
How do I verify the value of $\lambda$?
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Ensure that the coefficient of $x^2$ and $y^2$ are identical after multiplication. Here, $(1+3\lambda) = 2$ lead to $2 = 2$.
8
Is $ab$ the product of coordinates?
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Yes, $a$ is the x-coordinate of the centre and $b$ is the y-coordinate.
9
How much time should this take?
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Using the $\lambda$ method, this problem should take about 3 minutes.
10
What if there is an $xy$ term?
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If the conics had $xy$ terms, you would need to choose $\lambda$ such that the resulting $xy$ coefficient is zero to form a circle.