Mastering the equations of circles is a fundamental skill for competitive exams like JEE. This module focuses on a specific scenario: finding the equation of a circle that passes through two given points. We'll explore the underlying principles and methods to solve these problems efficiently.

Recall the standard form of a circle's equation: $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius. The general form is $x^2 + y^2 + 2gx + 2fy + c = 0$, where the center is $(-g, -f)$ and the radius is $\sqrt{g^2 + f^2 - c}$. Understanding these forms is crucial for deriving specific circle equations.

When a circle passes through two points, say A and B, the line segment AB can be considered a chord of that circle. A common method to find the circle's equation involves using the concept of the chord of contact. If we consider a point P from which tangents are drawn to the circle, the line joining the points of contact is the chord of contact. However, for a circle passing through two points, we can think of the line segment connecting these two points as a chord.

Let the two given points be $(x_1, y_1)$ and $(x_2, y_2)$. We can assume the general equation of the circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. Since the circle passes through $(x_1, y_1)$ and $(x_2, y_2)$, these points must satisfy the equation. This gives us two linear equations in terms of $g$, $f$, and $c$:

1. $x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c = 0$
2. $x_2^2 + y_2^2 + 2gx_2 + 2fy_2 + c = 0$

We need one more condition to uniquely determine $g$, $f$, and $c$. This condition might be the center lying on a specific line, the radius being a certain value, or another point the circle passes through.

If the two given points $(x_1, y_1)$ and $(x_2, y_2)$ are the extremities of a diameter of the circle, then the equation of the circle is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$. This is a direct and efficient method when the diameter endpoints are provided.

Let the equation of the line passing through the two points $(x_1, y_1)$ and $(x_2, y_2)$ be $L=0$. Let $S=0$ be the equation of any circle passing through these two points. The general equation of a circle passing through the intersection of $L=0$ and $S=0$ is $S + \lambda L = 0$. To use this, we need a base circle $S=0$ that passes through $(x_1, y_1)$ and $(x_2, y_2)$. A common choice for $S=0$ is the circle with the segment joining $(x_1, y_1)$ and $(x_2, y_2)$ as its diameter: $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$. Then, the family of circles is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) + \lambda (\text{equation of line AB}) = 0$. We then use any additional condition to find $\lambda$.

Find the equation of the circle passing through points (1, 2) and (3, 4) and whose center lies on the line $x + y = 5$.

Let the equation be $x^2 + y^2 + 2gx + 2fy + c = 0$. Substituting (1, 2): $1 + 4 + 2g + 4f + c = 0 \implies 2g + 4f + c = -5$ (1) Substituting (3, 4): $9 + 16 + 6g + 8f + c = 0 \implies 6g + 8f + c = -25$ (2) Subtracting (1) from (2): $4g + 4f = -20 \implies g + f = -5$ (3) Center is $(-g, -f)$. Since it lies on $x + y = 5$, we have $-g - f = 5$, which is $g + f = -5$. This is the same as (3), meaning the center condition is consistent with the points. We still need one more condition to find $g, f, c$. Let's assume the question implies these are the only two points given and we need to find a circle. If we assume these are diameter endpoints, the equation is $(x-1)(x-3) + (y-2)(y-4) = 0$, which simplifies to $x^2 - 4x + 3 + y^2 - 6y + 8 = 0$, or $x^2 + y^2 - 4x - 6y + 11 = 0$. Here, $g=-2, f=-3, c=11$. Center is $(2, 3)$. $2+3=5$, so it lies on $x+y=5$. This is a valid circle.

Line through (1, 2) and (3, 4): Slope $m = (4-2)/(3-1) = 2/2 = 1$. Equation: $y - 2 = 1(x - 1) \implies x - y + 1 = 0$. Let $L = x - y + 1$. Circle with diameter endpoints (1, 2) and (3, 4): $S = (x-1)(x-3) + (y-2)(y-4) = x^2 - 4x + 3 + y^2 - 6y + 8 = x^2 + y^2 - 4x - 6y + 11 = 0$. Family of circles: $S + \lambda L = 0 \implies x^2 + y^2 - 4x - 6y + 11 + \lambda(x - y + 1) = 0$. $x^2 + y^2 + (-4+\lambda)x + (-6-\lambda)y + (11+\lambda) = 0$. Center is $(-\frac{-4+\lambda}{2}, -\frac{-6-\lambda}{2}) = (\frac{4-\lambda}{2}, \frac{6+\lambda}{2})$. Center lies on $x+y=5$: $\frac{4-\lambda}{2} + \frac{6+\lambda}{2} = 5 \implies \frac{10}{2} = 5 \implies 5 = 5$. This means any value of $\lambda$ will satisfy the center condition if the base circle already satisfies it. This indicates that the problem might be underspecified or that the diameter form is the intended solution if no other constraints are given.

When faced with problems involving circles passing through two points:

1. Identify if the points are diameter endpoints. If yes, use the diameter form directly.
2. If not diameter endpoints, consider the family of circles approach ($S + \lambda L = 0$) or the general equation approach ($x^2 + y^2 + 2gx + 2fy + c = 0$) and use the given conditions to form equations.
3. Always check for consistency and ensure you have enough independent conditions to solve for the unknowns ($g, f, c$ or $\lambda$).