If m_i, 1/m_i, m_i 0, i = 1, 2, 3, 4 are four distinct points on a circle, then show that m₁ m₂ m₃

Q: Why do we assume the points lie on $xy=1$?

The problem defines the points as $(m_i, \frac{1}{m_i})$. If you multiply the x-coordinate ($m_i$) by the y-coordinate ($\frac{1}{m_i}$), you get $m_i \times \frac{1}{m_i} = 1$. This is the definition of the rectangular hyperbola $xy=1$. Any point whose coordinates satisfy this product must lie on this curve.

Q: Why do we multiply the equation by $m^2$?

When we substitute $x=m$ and $y=\frac{1}{m}$ into the circle equation $x^2 + y^2 + 2gx + 2fy + c = 0$, we get $m^2 + \frac{1}{m^2} + 2gm + \frac{2f}{m} + c = 0$. This equation has $m^2$ in the denominator. Multiplying by $m^2$ clears the fractions and transforms it into a standard quartic polynomial, which allows us to use Vieta's formulas easily.

Q: Can we use Vieta's formulas directly on the quartic equation?

Yes, absolutely! Once you have the quartic equation in the form $m^4 + 2gm^3 + cm^2 + 2fm + 1 = 0$, the product of the roots is given by the constant term divided by the leading coefficient. Since both are 1, the product is 1. It is a beautiful application of the theory of equations.

Question

If $m_i, \frac{1}{m_i}, m_i > 0, i = 1, 2, 3, 4$ are four distinct points on a circle, then show that $m_1 m_2 m_3 m_4 = 1$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, today we are going to witness a beautiful convergence. We are looking at four distinct points, each defined by the coordinates $(m_i, \frac{1}{m_i})$.

At first glance, these might seem like arbitrary points, but let's look closer. If you take the $x$-coordinate $m_i$ and multiply it by the $y$-coordinate $\frac{1}{m_i}$, you get exactly $1$.

This is the signature of a rectangular hyperbola, $xy=1$. Imagine this curve plotted on your coordinate plane; it is a classic, symmetric shape. Now, the problem tells us that these four points also lie on a circle. This is the crux of our journey.

### The General Tool

To handle this, we need a robust mathematical framework. We invoke the general equation of a circle:

$$x^2 + y^2 + 2gx + 2fy + c = 0$$

This equation represents any circle in the Cartesian plane. Since our four points $(m_i, \frac{1}{m_i})$ lie on this circle, they must satisfy this equation perfectly. This is our bridge between geometry and algebra.

We substitute $x = m$ and $y = \frac{1}{m}$ into our circle equation. The result is:

$$m^2 + \left(\frac{1}{m}\right)^2 + 2g(m) + 2f\left(\frac{1}{m}\right) + c = 0$$

### The Quartic Transformation

Now, I know what you are thinking: that $\frac{1}{m^2}$ term looks troublesome. But in mathematics, as in life, we often need to clear the path before we can see the destination.

To eliminate the fractions, we multiply the entire equation by $m^2$. This gives us:

$$m^4 + 1 + 2gm^3 + 2fm + cm^2 = 0$$

Let's rearrange this into a standard, descending-power format:

$$m^4 + 2gm^3 + cm^2 + 2fm + 1 = 0$$

This is a quartic equation—a fourth-degree polynomial. Because we have four distinct points of intersection, this equation must have exactly four roots: $m_1, m_2, m_3,$ and $m_4$.

### The Vieta's Finale

This is where the magic happens. According to the theory of equations, specifically Vieta's formulas, for any polynomial of the form $ax^4 + bx^3 + cx^2 + dx + e = 0$, the product of the roots is given by $\frac{e}{a}$.

In our specific quartic equation, the leading coefficient $a$ is $1$, and the constant term $e$ is also $1$. Therefore, the product of the roots is simply:

$$\frac{1}{1} = 1$$

Thus, **$m_1 m_2 m_3 m_4 = 1$**.

It is a stunning result, isn't it? We started with a geometric intersection and arrived at a profound algebraic truth. Keep this elegance in mind as you solve more problems; often, the most complex-looking geometry hides a simple, beautiful algebraic soul.

Animated Solution for Mathematics - Circles: If $m_{i}, \frac{1}{m _{i}}, m_{i} > 0, i = 1, 2, 3, 4$ are four distinct points on a circle, then show that $m_{1} m_{2} m_{3} m_{4} = 1$ .

Visualized Solution

Points on a Hyperbola

Intersection with a Circle

General Equation of a Circle

Substituting the Coordinates

The Substituted Equation

Eliminating Fractions

Expanding the Terms

Forming the Quartic Equation

Roots of the Quartic

Theory of Equations

Applying Vieta's Formula

The Final Result

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The General Tool

The Quartic Transformation

The Vieta's Finale

Similar Questions

If the four distinct points $(4, 6), (- 1, 5), (0, 0)$ and $(k, 3 k)$ lie on a circle of radius $r$ , then $10 k + r^{2}$ is equal to

Let $C$ be any circle with centre $(0, 2)$ . Prove that at the most two rational points can be there on $C$ . (A rational point is a point both of whose coordinates are rational numbers.)

Four distinct points $(2 k, 3 k), (1, 0), (0, 1)$ and $(0, 0)$ lie on a circle for $k$ equal to :

If $P$ is a point on the circle $x^{2} + y^{2} = 4$ , $Q$ is a point on the straight line $5 x + y + 2 = 0$ and $x - y + 1 = 0$ is the perpendicular bisector of $P Q$ , then 13 times the sum of abscissa of all such points $P$ is .........

Let $P$ and $Q$ be two distinct points on a circle which has center at $C (2, 3)$ and which passes through origin $O$ . If $O C$ is perpendicular to both the line segments $C P$ and $C Q$ , then the set ${P, Q}$ is equal to

Through a fixed point $(h, k)$ secants are drawn to the circle $x^{2} + y^{2} = r^{2}$ . Show that the locus of the mid-points of the secants intercepted by the circle is $x^{2} + y^{2} = h x + k y$ .

Let $C_{1}$ be the circle in the third quadrant of radius 3, that touches both coordinate axes. Let $C_{2}$ be the circle with centre $(1, 3)$ that touches $C_{1}$ externally at the point $(α, β)$ . If $(β - α)^{2} = \frac{m}{n}, gcd (m, n) = 1$ , then $m + n$ is equal to :

Three distinct points $A, B$ and $C$ are given in the 2-dimensional coordinates plane such that the ratio of the distance of any one of them from the point $(1, 0)$ to the distance from the point $(- 1, 0)$ is equal to $\frac{1}{3}$ . Then the circumcentre of the triangle $A B C$ is at the point

Animated Solution for Mathematics - Circles: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If mi​,mi​1​,mi​>0,i=1,2,3,4 are four distinct points on a circle, then show that m1​m2​m3​m4​=1.