If A and B are points in the plane such that PA/PB = k (constant) for all P on a given circle, then

Q: Why do we have to square the equation? Can't we just solve it with the square roots?

Technically, you could, but it would be a nightmare! Squaring is a standard algebraic maneuver to eliminate radicals. It keeps the equation in a polynomial form, which is much easier to analyze for circle properties. If you keep the square roots, you will struggle to isolate the $x^2$ and $y^2$ terms effectively.

Q: What if $k$ is negative?

Great question! Remember that $k$ represents a ratio of distances, and distances are always non-negative. Therefore, $k$ must be a positive real number. If $k$ were negative, the ratio would be undefined in a physical sense, so we only consider $k > 0$.

Q: Why does $k=1$ result in a straight line instead of a circle?

When $k=1$, the equation $PA = PB$ defines the locus of points equidistant from $A$ and $B$. Geometrically, this is the perpendicular bisector of the segment $AB$. Since a line has no curvature, it cannot be a circle. The circle of Apollonius only exists when the ratio is not unity.

Question

If $A$ and $B$ are points in the plane such that $PA/PB = k$ (constant) for all $P$ on a given circle, then the value of $k$ cannot be equal to .........

Sigma Percentile · Accepted Answer

### The Geometry of Balance: Unveiling the Circle of Apollonius

Welcome, future engineer. Today, we are not just solving an equation; we are uncovering a beautiful geometric truth. We are exploring the Circle of Apollonius.

Imagine you are standing on a vast, empty plane with two fixed anchors, point $A$ and point $B$. Now, imagine a point $P$ that is dancing around this plane, bound by a secret rule: the ratio of its distance from $A$ to its distance from $B$ is always a constant, $k$.

That is, the condition is defined as:
$$\frac{PA}{PB} = k$$
This ratio is the heartbeat of our problem.

### The Algebraic Trap

I know what you are thinking. Let's just plug in the distance formula and start calculating. But wait! If you write the expression as:
$$\sqrt{(x-x_1)^2 + (y-y_1)^2} = k \sqrt{(x-x_2)^2 + (y-y_2)^2}$$
you are walking into a trap.

Those square roots are not just symbols; they are algebraic landmines. They make the equation look terrifying and impossible to simplify.

The secret, the masterstroke, is to square both sides immediately. By transforming the condition into $PA^2 = k^2 PB^2$, we strip away the radicals. Now, we are dealing with a clean, manageable polynomial equation:
$$(x - x_1)^2 + (y - y_1)^2 = k^2 [(x - x_2)^2 + (y - y_2)^2]$$

### The Expansion and the Revelation

Now, take a deep breath and expand both sides. You will see terms like $x^2$, $y^2$, linear terms, and constants.

It might look like a mess, but look closer at the coefficients of $x^2$ and $y^2$. When you bring everything to one side, the coefficient of $x^2$ becomes $(1 - k^2)$ and the coefficient of $y^2$ also becomes $(1 - k^2)$.

This is the moment of truth. For an equation to represent a circle, we need the coefficients of $x^2$ and $y^2$ to be equal and non-zero.

If they are zero, the circle collapses. This happens exactly when $1 - k^2 = 0$, or $k = 1$.

When $k=1$, the equation simplifies to a linear equation, which is the **perpendicular bisector** of $AB$. It is a straight line, not a circle.

Thus, for the locus to be a circle, **$k$ cannot be $1$**. You have just mastered the Circle of Apollonius. Keep this intuition, and no geometry problem will ever intimidate you again.

Animated Solution for Mathematics - Circles: If $A$ and $B$ are points in the plane such that $P A / P B = k$ (constant) for all $P$ on a given circle, then the value of $k$ cannot be equal to .........

Enter Numerical Value:

Visualized Solution

Setting the Stage

The Given Condition

Squaring the Condition

Applying Distance Formula

Expanding the Terms

Grouping Second Degree Terms

Condition for a Circle

Analyzing $k = 1$

The Locus when $k = 1$

Final Conclusion

The Sigma Insight: Standard and General Equation of a Circle

The Geometry of Balance

The Algebraic Trap

The Expansion and the Revelation

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