Question

If the equation $x^2 + y^2 -10x + 21 = 0$ has real roots $x = a$ and $y=\beta$ then

• A. $3\le x\le7$
• B. $3\le y\le7$
• C. $-2 \le y \le2$
• D. $-2\le x\le2$

Answer 1

Answer: (C)

$-2 \le y \le2$

Answer 2

$x^{2}-10x+\left(y^{2}+21\right)=0$
for real roots of $x, D \ge0$
$100-4\left(y^{2}+21\right)\ge0$
$\Rightarrow y^{2}\le4$
$\Rightarrow -2\le y\le2 \left(C\right)$
also, $y^{2}=-x^{2}+10x-21$
for real roots of $y,$
$-x^{2}+10x-21 \ge0$
$\Rightarrow \left(x-7\right)\left(x-3\right)\le0$
$3\le x\le7 \left(A\right)$