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If the equation \(x^{2}+y^{2}-10 x+21=0\) has real roots \(x=a\) and \(y=\beta\) then
(A) \(3 \leq x \leq 7\)
(B) \(3 \leq y \leq 7\)
(C) \(-2 \leq y \leq 2\)
(D) \(-2 \leq x \leq 2\)

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Ans : (A, C)
Hint: \(\quad x^{2}-10 x+\left(y^{2}+21\right)=0\)
for real roots of \(x, D \geq 0\)
\(100-4\left(y^{2}+21\right) \geq 0\)
\(\Rightarrow \mathrm{y}^{2} \leq 4\)
\(\Rightarrow-2 \leq y \leq 2\) (C)
also, \(y^{2}=-x^{2}+10 x-21\)
for real roots of \(y\),
$$
\begin{aligned}
&-x^{2}+10 x-21 \geq 0 \\
\Rightarrow &(x-7)(x-3) \leq 0 \\
& 3 \leq x \leq 7 \text { (A) }
\end{aligned}
$$
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