If the equation x^{2}+y^{2}-10x+21=0 has real | vedclass

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(C) Given equation: $x^{2} - 10x + y^{2} + 21 = 0$.For $x$ to have real roots,the discriminant $D$ must be greater than or equal to $0$.Treating the e

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$-2 \leq y \leq 2$

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(C) Given equation: $x^{2} - 10x + y^{2} + 21 = 0$.For $x$ to have real roots,the discriminant $D$ must be greater than or equal to $0$.Treating the equation as a quadratic in $x$: $x^{2} - 10x + (y^{2} + 21) = 0$.$D = (-10)^{2} - 4(1)(y^{2} + 21) \geq 0$.$100 - 4y^{2} - 84 \geq 0$ $\Rightarrow 16 - 4y^{2} \geq 0$ $\Rightarrow y^{2} \leq 4$.Thus,$-2 \leq y \leq 2$.Similarly,for $y$ to have real roots,treat the equation as a quadratic in $y$: $y^{2} + (x^{2} - 10x + 21) = 0$.Since $y^{2} = -x^{2} + 10x - 21$,for $y$ to be real,$y^{2} \geq 0$.$-x^{2} + 10x - 21 \geq 0 \Rightarrow x^{2} - 10x + 21 \leq 0$.$(x-7)(x-3) \leq 0 \Rightarrow 3 \leq x \leq 7$.

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

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