Let and , where is a real number. If the largest possible value of is p, then the circle , intersects the co-ordinate axes at [2026]

Question

Let $A = [\begin{matrix} 1 & 2 & 7 \\ 4 & - 2 & 8 \\ 3 & 8 & - 7 \end{matrix}]$ and $\det (A - α I) = 0$ , where $α$ is a real number. If the largest possible value of $α$ is p, then the circle ${(x - p)}^{2} + {(y - 2 p)}^{2} = 320$ , intersects the co-ordinate axes at [2026]

Smart Achievers · Accepted Answer

(3)

$A = [\begin{matrix} 1 & 2 & 7 \\ 4 & - 2 & 8 \\ 3 & 8 & - 7 \end{matrix}]$

Now, $| A - α I | = 0$

$α$

$\Rightarrow (1 - α) {α^{2} + 9 α + 14 - 64} - 2 {- 4 α - 28 - 24} + 7 {32 + 6 + 3 α} = 0$

$\Rightarrow α^{2} + 9 α - 50 - α^{3} - 9 α^{2} + 50 α + 8 α + 104 + 266 + 21 α = 0$

$\Rightarrow α^{2} + 8 α^{2} - 88 α - 320 = 0$

$\Rightarrow α^{2} (α = 8) + 16 α (α - 8) + 40 (α - 8) = 0$

$\Rightarrow (α = 8) + (α^{2} + 16 α + 40) = 0$

$\Rightarrow α = 8, α = \frac{- 16 \pm \sqrt{96}}{2}$

$\Rightarrow α = - 8 \pm \sqrt{24}$

Largest possible value of $α$ is 8.

$∴ p = 8$

For y = 0

${(x - 8)}^{2} = {(y - 16)}^{2} = 320$

$\Rightarrow x - 8 = \pm 8 \Rightarrow x = 16, 0$

For x = 0

${(y - 16)}^{2} = 320 - 64$

$\Rightarrow {(y - 16)}^{2} = 256 \Rightarrow y - 16 = \pm 16$

y = 32, 0

So, points are (0, 32), (0, 0), (16, 0).

Solution

Q.
Let $A = [\begin{matrix} 1 & 2 & 7 \\ 4 & - 2 & 8 \\ 3 & 8 & - 7 \end{matrix}]$ and $\det (A - α I) = 0$ , where $α$ is a real number. If the largest possible value of $α$ is p, then the circle ${(x - p)}^{2} + {(y - 2 p)}^{2} = 320$ , intersects the co-ordinate axes at [2026]

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