Question
Let $A = \begin{bmatrix} 1 & 2 & 7 \\ 4 & - 2 & 8 \\ 3 & 8 & - 7 \end{bmatrix}$ and $det(A - \alpha I) = 0$,
where $\alpha$ is a real number. If the largest possible value of $\alpha$ is $p$, then the circle $(x - p)^{2} + (y - 2p)^{2} =$ 320, intersects the co-ordinate axes at
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Solution
Sol. [Key Concept: Expand the characteristic cubic $det(A - \alpha\ I) = 0$ for the largest root $p$; the circle meets an axis where $x = 0$ or $y = 0$.]
$$\left| \begin{matrix} 1 - \alpha & 2 & 7 \\ 4 & - 2 - \alpha & 8 \\ 3 & 8 & - 7 - \alpha \end{matrix} \right| = 0$$
$$\begin{matrix} \Rightarrow & \ (1 - \alpha)\lbrack(\alpha + 2)(\alpha + 7) - 64\rbrack - 2\lbrack - 28 - 4\alpha - \\ & 24\rbrack + 7\lbrack 32 + 6 + 3\alpha\rbrack = 0 \\ \Rightarrow & \alpha^{3} + 8\alpha^{2} - 88\alpha - 320 = 0 \\ & \ (\alpha - 8)\left( \alpha^{2} + 16\alpha + 40 \right) = 0 \\ & \alpha = 8,\alpha = - 8 \pm 2\sqrt{6} \end{matrix}$$
So $p = - 8$
$$\Rightarrow (x - 8)^{2} + (y - 16)^{2} = 320 = 8^{2} + 16^{2} $$put $y = 0 \Rightarrow x = 16,0$
put $x = 0 \Rightarrow y = 32,0$
$$\Rightarrow (16,0),(0,32),(0,0) $$⇒ 3 points of intersection
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