If m_i, 1/m_i, m_i > 0, i = 1, 2, 3, 4...

MathematicsStandard and General Equation of a CircleJEE Advanced 1989Moderate

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 (Hindi)

Visualizing the Problem

Consider four distinct points $(m_{i}, \frac{1}{m _{i}})$ for $i = 1, 2, 3, 4$ .
These points lie on the rectangular hyperbola $x y = 1$ .
They also lie on a circle whose equation we need to define.

The General Circle Equation

Let the general equation of the circle be:
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
where $g, f, c$ are constants.

Defining the Points

The points are of the form $(m, \frac{1}{m})$ .
Since these points lie on the circle, they must satisfy its equation.
We have four such values: $m_{1}, m_{2}, m_{3}, m_{4}$ .

Substitution Step

Substitute $x = m$ and $y = \frac{1}{m}$ into the circle equation:
$m^{2} + (\frac{1}{m})^{2} + 2 g (m) + 2 f (\frac{1}{m}) + c = 0$
This simplifies to: $m^{2} + \frac{1}{m ^{2}} + 2 g m + \frac{2 f}{m} + c = 0$

Clearing the Denominators

To eliminate the fractions, multiply the entire equation by $m^{2}$ :
$m^{2} (m^{2} + \frac{1}{m ^{2}} + 2 g m + \frac{2 f}{m} + c) = 0$
Expanding this gives: $m^{4} + 1 + 2 g m^{3} + 2 f m + c m^{2} = 0$

The Quartic Equation Form

Rearrange the terms in descending order of $m$ :
$m^{4} + 2 g m^{3} + c m^{2} + 2 f m + 1 = 0$
This is a quartic equation in $m$ .

Applying Vieta's Formulas

For a quartic equation $a x^{4} + b x^{3} + c x^{2} + d x + e = 0$ , the product of roots is given by $\frac{e}{a}$ .
In our equation: $a = 1$ and $e = 1$ .
Therefore, the product of roots $m_{1} m_{2} m_{3} m_{4} = \frac{1}{1}$ .

Final Result

The product of the four distinct values is:
$m_{1} m_{2} m_{3} m_{4} = 1$
Hence proved.

The Way Forward

Key Takeaway: The intersection of a circle and a rectangular hyperbola leads to a quartic equation where the product of the x-coordinates is related to the constant term.
Next Challenge: What if the points were on a parabola $y^{2} = 4 a x$ ? How would the product of the ordinates behave?

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.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.

Visualized Solution (Hindi)

Visualizing the Problem

The General Circle Equation

Defining the Points

Substitution Step

Clearing the Denominators

The Quartic Equation Form

Applying Vieta's Formulas

Final Result

The Way Forward

Conceptually Similar Problems

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A,B,C,D are any four points in space, prove that ∣AB×CD+BC×AD+CA×BD∣=4(area of triangle ABC).

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 1,a1​,a2​,…,an−1​ are the n roots of unity, then show that (1−a1​)(1−a2​)(1−a3​)…(1−an−1​)=n.

.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.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the circle x2+y2=a2 intersects the hyperbola xy=c2 in four points P(x1​,y1​),Q(x2​,y2​),R(x3​,y3​),S(x4​,y4​), then

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If $1, a_{1}, a_{2}, \dots, a_{n - 1}$ are the $n$ roots of unity, then show that $(1 - a_{1}) (1 - a_{2}) (1 - a_{3}) \dots (1 - a_{n - 1}) = n$ .