Let C be a circle passing through the points A(2, -1) and B(3, 4). The line segment AB is not a diam

Q: Wait, why do we reject the first solution for y?

Great catch! When we solve the quadratic equation, we get two possible values for y. If we plug y = 1.5 back into our line equation, we get x = 2.5. This point (2.5, 1.5) is exactly the midpoint of the chord AB. If the center of the circle were the midpoint of the chord, then AB would be a diameter. The problem explicitly states that AB is *not* a diameter, so we must reject this case.

Q: Why can't we just use the distance formula from the start?

You could try, but you would end up with two variables (h and k) and only one equation (the distance from the center to A equals the distance from the center to B). That is not enough to solve for the center. We need the second constraint provided by the locus circle to create a system of equations.

Q: Is there a faster way to find the perpendicular bisector?

The method we used is the most reliable. You could use the property that the distance from the center to A equals the distance from the center to B, i.e., (h-2)^2 + (k+1)^2 = (h-3)^2 + (k-4)^2. If you expand this, you will get the exact same linear equation for the perpendicular bisector. Both methods are valid, but the slope-point method is often less prone to algebraic errors.

Question

Let $C$ be a circle passing through the points $A(2, -1)$ and $B(3, 4)$. The line segment $AB$ is not a diameter of $C$. If $r$ is the radius of $C$ and its centre lies on the circle $(x - 5)^2 + (y - 1)^2 = \frac{13}{2}$, then $r^2$ is equal to :

Options:

(A) 32

(B) $\frac{65}{2}$

(C) $\frac{61}{2}$

(D) 30

Sigma Percentile · Accepted Answer

### The Geometry of Constraints: Unlocking the Circle

Imagine you are standing on a coordinate plane, looking at two points: $A(2, -1)$ and $B(3, 4)$. A circle passes through both of them. This is not just a random arrangement; it is a geometric dance.

Because the circle passes through $A$ and $B$, the segment $AB$ is a chord. The center of any circle must lie on the perpendicular bisector of its chords. This is the first thread we must pull to unravel the mystery of the circle's center.

### Phase 1: The Perpendicular Bisector

To find this bisector, we need two things: a point on the line and the slope of the line. The point is the midpoint $M$ of the chord $AB$.

Calculating the midpoint:
$$M = \left( \frac{2 + 3}{2}, \frac{-1 + 4}{2} \right) = (2.5, 1.5)$$

Next, we find the slope of the chord $AB$ using the slope formula $m_1 = \frac{y_2 - y_1}{x_2 - x_1}$:
$$m_1 = \frac{4 - (-1)}{3 - 2} = 5$$

Since our bisector is perpendicular to the chord, its slope $m_2$ must be the negative reciprocal of $m_1$. Thus, $m_2 = -\frac{1}{5}$.

Using the point-slope form $y - y_M = m_2(x - x_M)$, we write:
$$y - 1.5 = -\frac{1}{5}(x - 2.5)$$

Simplifying this, we get $5(y - 1.5) = -(x - 2.5)$, which leads us to the linear equation:
$$x + 5y = 10$$
This line is the home of our circle's center.

### Phase 2: The Locus and the Intersection

The center $(h, k)$ also lies on the line $x = 10 - 5y$. We are given that the center must satisfy the constraint of the circle equation:
$$(x - 5)^2 + (y - 1)^2 = 6.5$$

By substituting the expression for $x$ into the circle equation, we get:
$$(10 - 5y - 5)^2 + (y - 1)^2 = 6.5$$
$$(5 - 5y)^2 + (y - 1)^2 = 6.5$$

Factoring out the 5, we get $25(1 - y)^2 + (y - 1)^2 = 6.5$. Since $(1 - y)^2$ is equivalent to $(y - 1)^2$, this simplifies to:
$$26(y - 1)^2 = 6.5$$

Solving for $y$:
$$(y - 1)^2 = \frac{6.5}{26} = 0.25 = \frac{1}{4}$$
This yields two potential values: $y = 1.5$ or $y = 0.5$.

### Phase 3: The Diameter Trap

Here is where many students stumble. We have two possible centers.

If $y = 1.5$, then $x = 10 - 5(1.5) = 2.5$. This point $(2.5, 1.5)$ is our midpoint $M$. If the center were the midpoint, $AB$ would be a diameter, which the problem forbids.

We must reject this case. The only valid center is $O(7.5, 0.5)$.

Finally, we calculate the radius squared, $r^2$, using the distance formula from the center $O(7.5, 0.5)$ to point $A(2, -1)$:
$$r^2 = (7.5 - 2)^2 + (0.5 - (-1))^2$$
$$r^2 = (5.5)^2 + (1.5)^2 = 30.25 + 2.25 = 32.5$$

Expressed as a fraction, the final result is:
**$r^2 = \frac{65}{2}$**

Select Answer:

Visualized Solution

Visualizing the Points $A$ and $B$

The Perpendicular Bisector Property

Finding the Midpoint of $A B$

Calculating the Slope of $A B$

Slope of the Perpendicular Bisector

Equation of the Perpendicular Bisector

The Second Constraint: The Locus Circle

Substitution and Algebra

Simplifying the Quadratic Equation

Solving for $y$

JEE Trap: The Diameter Constraint

Finding the Valid Center

Calculating $r^{2}$ using Distance Formula

Final Computation

Summary and Conclusion

The Sigma Insight: Standard and General Equation of a Circle

The Geometry of Constraints

Phase 1

Phase 2

Phase 3

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