MathematicsStandard and General Equation of a CircleJEE Advanced 1989Moderate
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Visualized Solution (English)

Visualizing the Problem

  • Consider four distinct points for .
  • These points lie on the rectangular hyperbola .
  • They also lie on a circle whose equation we need to define.

The General Circle Equation

  • Let the general equation of the circle be:
  • where are constants.

Defining the Points

  • The points are of the form .
  • Since these points lie on the circle, they must satisfy its equation.
  • We have four such values: .

Substitution Step

  • Substitute and into the circle equation:
  • This simplifies to:

Clearing the Denominators

  • To eliminate the fractions, multiply the entire equation by :
  • Expanding this gives:

The Quartic Equation Form

  • Rearrange the terms in descending order of :
  • This is a quartic equation in .

Applying Vieta's Formulas

  • For a quartic equation , the product of roots is given by .
  • In our equation: and .
  • Therefore, the product of roots .

Final Result

  • The product of the four distinct values is:
  • 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 ? How would the product of the ordinates behave?

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