Let C be any circle with centre (0,...

MathematicsStandard and General Equation of a CircleJEE Advanced 1997Moderate

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.)

Visualized Solution (English)

Visualizing the Circle and Center

Center of the circle: $C (0, 2)$
A rational point $(x, y)$ is one where $x, y \in Q$ .
The center is located on the $y$ -axis at an irrational height.

Defining the Circle Equation

Standard Equation: $(x - h)^{2} + (y - k)^{2} = r^{2}$
Substituting $(0, 2)$ : $x^{2} + (y - 2)^{2} = r^{2}$
Expanding: $x^{2} + y^{2} - 22 y + 2 = r^{2}$

Assuming Two Rational Points

Let $P_{1} (x_{1}, y_{1})$ and $P_{2} (x_{2}, y_{2})$ be two rational points.
Eq 1: $x_{1}^{2} + y_{1}^{2} - 22 y_{1} + 2 = r^{2}$
Eq 2: $x_{2}^{2} + y_{2}^{2} - 22 y_{2} + 2 = r^{2}$

Eliminating the Radius

Subtracting Eq 2 from Eq 1:
$(x_{1}^{2} + y_{1}^{2}) - (x_{2}^{2} + y_{2}^{2}) = 22 (y_{1} - y_{2})$

The Rationality Constraint

LHS: $(x_{1}^{2} + y_{1}^{2}) - (x_{2}^{2} + y_{2}^{2}) \in Q$
RHS: $22 (y_{1} - y_{2})$
For $Rational = Irrational \times Rational$ , the rational multiplier must be zero.
$⟹ y_{1} - y_{2} = 0 ⟹ y_{1} = y_{2}$

Solving for x-coordinates

Since $y_{1} = y_{2}$ , substituting back into the circle equation:
$x_{1}^{2} + (y_{1} - 2)^{2} = x_{2}^{2} + (y_{1} - 2)^{2}$
$⟹ x_{1}^{2} = x_{2}^{2} ⟹ x_{1} = \pm x_{2}$
This gives at most two distinct points: $(x_{1}, y_{1})$ and $(- x_{1}, y_{1})$ .

Final Conclusion

Key Takeaway: The irrational y-coordinate of the center forces all rational points to have the same y-value.
Geometric Fact: A circle can intersect a horizontal line at most at two points.
Final Result: At most two rational points can exist on $C$ .

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$C_{1}$ and $C_{2}$ are two concentric circles, the radius of $C_{2}$ being twice that of $C_{1}$ . From a point $P$ on $C_{2}$ , tangents $P A$ and $P B$ are drawn to $C_{1}$ . Prove that the centroid of the triangle $P A B$ lies on $C_{1}$ .

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Let $R S$ be the diameter of the circle $x^{2} + y^{2} = 1$ , where $S$ is the point $(1, 0)$ . Let $P$ be a variable point (other than $R$ and $S$ ) on the circle and tangents to the circle at $S$ and $P$ meet at the point $Q$ . The normal to the circle at $P$ intersects a line drawn through $Q$ parallel to $R S$ at point $E$ . Then the locus of $E$ passes through the point(s)

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Visualized Solution (English)

Visualizing the Circle and Center

Defining the Circle Equation

Assuming Two Rational Points

Eliminating the Radius

The Rationality Constraint

Solving for x-coordinates

Final Conclusion

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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 $m_{1} m_{2} m_{3} m_{4} = 1$ .

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If the vertices $P, Q, R$ of a triangle $P QR$ are rational points, which of the following points of the triangle $P QR$ is (are) always rational point(s)?

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 $x^{2} + y^{2} = h x + k y$ .

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