If the locus of the mid-point of the line segment from the point (3,2) to a point on the circle, x²

Q: Wait, why do we use $(h, k)$ instead of $(x, y)$ initially?

That is a fantastic question! We use $(h, k)$ to represent the coordinates of the midpoint $Q$ to distinguish it from the variable point $P(x, y)$ on the circle. It helps us keep our algebra clean and prevents us from confusing the moving point on the circle with the point whose locus we are trying to find. Once we derive the equation in terms of $h$ and $k$, we can simply replace them with $x$ and $y$ to write the final locus equation.

Q: Can we use geometry instead of algebra to solve this?

Absolutely! You can use the concept of homothety or scaling. The midpoint $Q$ is essentially the image of point $P$ under a transformation that scales the plane by a factor of $1/2$ with respect to the fixed point $A$. Since the original shape is a circle, the image under this scaling transformation will also be a circle, and its radius will be scaled by the same factor of $1/2$. It is a beautiful way to see the result without heavy calculation!

Q: Why does the radius become half of the original?

Think of the midpoint formula as an average. When you take the midpoint of a segment connecting a fixed point to a point on a circle, you are essentially shrinking the distance from the fixed point to the circle by half. Because the entire circle is being 'compressed' towards the fixed point by a factor of $1/2$, all linear dimensions of the circle—including its radius—are scaled by that same factor. That is why the new radius is $1/2$ of the original radius of $1$.

Question

If the locus of the mid-point of the line segment from the point $(3,2)$ to a point on the circle, $x^2 + y^2 = 1$ is a circle of the radius $r$, then $r$ is equal to :

Options:

(A) $\frac{1}{4}$

(B) $\frac{1}{2}$

(C) 1

(D) $\frac{1}{3}$

Sigma Percentile · Accepted Answer

### The Locus of the Midpoint: A Journey into Geometric Elegance

My dear student, welcome to a beautiful exploration of coordinate geometry. Today, we are not just solving a problem; we are uncovering the hidden symmetry of a locus.

Imagine you are standing on a vast, flat plane. At the origin, there is a perfect circle with a radius of $1$. Far away, at the coordinates $(3, 2)$, sits a fixed anchor point, $A$.

Now, imagine a point $P$ dancing along the circumference of that circle. As $P$ moves, we are interested in the path traced by the midpoint $Q$ of the line segment $AP$. What does this path look like? Let us find out together.

### The Setup: The Anchor and the Dancer

First, let us define our players. The circle is defined by the equation $x^2 + y^2 = 1$. Our fixed point is $A(3, 2)$.

Our dancer, point $P$, has coordinates $(x_1, y_1)$. Because $P$ is constrained to the circle, it must satisfy the equation $x_1^2 + y_1^2 = 1$. This is our fundamental constraint—the rule that governs the dancer's movement.

Now, we introduce our midpoint $Q(h, k)$. By the definition of a midpoint, $h$ is the average of the $x$-coordinates of $A$ and $P$, and $k$ is the average of the $y$-coordinates. Thus, we have:

$$h = \frac{x_1 + 3}{2}, \quad k = \frac{y_1 + 2}{2}$$

### The Algebraic Bridge

To find the locus of $Q$, we need an equation that relates $h$ and $k$ directly, without the interference of $x_1$ and $y_1$. We must bridge the gap between the dancer's position and the midpoint's path.

We rearrange our midpoint equations to isolate $x_1$ and $y_1$:

$$x_1 = 2h - 3, \quad y_1 = 2k - 2$$

This is the key! We have expressed the dancer's position in terms of the midpoint's coordinates. Now, we substitute these into our constraint equation, $x_1^2 + y_1^2 = 1$. This gives us:

$$(2h - 3)^2 + (2k - 2)^2 = 1$$

### The Algebraic Alchemy

Do not let this equation intimidate you. It is just a circle in disguise. To reveal its true form, we need the coefficients of $h$ and $k$ to be $1$.

Let us factor out a $2$ from each bracket. Remember, when we factor a $2$ out of a squared term, it becomes a $4$:

$$4\left(h - \frac{3}{2}\right)^2 + 4(k - 1)^2 = 1$$

Now, divide the entire equation by $4$. We get:

$$\left(h - \frac{3}{2}\right)^2 + (k - 1)^2 = \frac{1}{4}$$

This is the equation of the path traced by $Q$.

### The Reveal

Finally, let us replace $(h, k)$ with $(x, y)$ to write the locus in standard form:

$$\left(x - \frac{3}{2}\right)^2 + (y - 1)^2 = \left(\frac{1}{2}\right)^2$$

Comparing this to the standard circle equation $(x - x_c)^2 + (y - y_c)^2 = r^2$, we can clearly see that the radius $r$ is $\frac{1}{2}$.

The **locus of the midpoint is another circle**, scaled down by exactly half. Isn't that elegant? The midpoint doesn't just wander aimlessly; it traces a perfect, smaller circle. I hope this journey has helped you see the beauty in the algebra.

Animated Solution for Mathematics - Circles: If the locus of the mid-point of the line segment from the point $(3, 2)$ to a point on the circle, $x^{2} + y^{2} = 1$ is a circle of the radius $r$ , then $r$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Setup

Defining Point $P$

Constraint for Point $P$

Introducing the Mid-point $Q$

Applying the Mid-point Formula

Isolating $x_{1}$ and $y_{1}$

Substituting into the Constraint

Factoring Out Constants

Standardizing the Locus Equation

Identifying the Radius

Final Conclusion

The Sigma Insight: Standard and General Equation of a Circle

The Locus of the Midpoint

The Setup

The Algebraic Bridge

The Algebraic Alchemy

The Reveal

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