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$S(x, y) = 0$ represents a circle. The equation $S(x, 2) = 0$ gives two identical solutions $x = 1$ and the equation $S(1, y) = 0$ gives two distinct solutions $y=0,2.$ Find the equation of the circle.
Option: 1
$x^{2}+y^{2}+2x-2y+1=0$

Option: 2
$x^{2}+y^{2}-2x+2y+1=0$

Option: 3
$x^{2}+y^{2}-2x-2y-1=0$

Option: 4
$x^{2}+y^{2}-2x-2y+1=0$

Answers (1)

best_answer

Equation of a circle in diametric form -

$\left ( x-x_{1} \right )\left ( x-x_{2} \right )+\left ( y-y_{1} \right )\left ( y-y_{2} \right )= 0$

- wherein

Where $A\left ( x_{1},y_{1} \right )\, and \:B \left ( x_{2},y_{2} \right )$ are the two diametric ends.

S(x, 2) = 0 given two identical solutions x = 1.

$\Rightarrow$ line y = 2 is a tangent to the circle S(x, y) = 0 at the point (1, 2) and S(1, y) = 0 gives two distinct solutions y = 0, 2

$\Rightarrow$ Line x = 1 cut the circle S(x, y) = 0 at points (1, 0) and (1, 2)

A(1, 2) and B(1, 0) are diametrically opposite points.

$\therefore$ equation of the circle is $\left ( x-1 \right )^{2}+y\left ( y-2 \right )=0$

$x^{2}+y^{2}-2x-2y+1=0$

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Suraj Bhandari

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