First slide

Theory of equations

Question

$x^{2} - xy + y^{2} - 4 x - 4 y + 16 = 0$ represents

Moderate

A

a point
B

a circle
C

a pair of straight lines
D

none of these

Solution

Given equation is $x^{2} - (y + 4) x + y^{2} - 4 y + 16 = 0$
Since x is real, we have
$D \geq 0$
$\Rightarrow (y + 4)^{2} - 4 (y^{2} - 4 y + 16) \geq 0$
or $- 3 y^{2} + 24 y - 48 \geq 0$
or $y^{2} - 8 y + 16 \leq 0$
or $(y - 4)^{2} \leq 0$
or y - 4 = 0
or y = 4
Since the equation is symmetric in r and y, we have x = 4 only.

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