Q. If each of the lines $5x + 8y = 13$ and $4x-y = 3$ contains a diameter of the circle $x^2+y^2 - 2(a^2 - 7a +11)$
$x-2 (a^2-6a + 6)y + b^3 +1 = 0$, then :

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$a = 5$ and $b ∉ \left(-1,1\right)$

$a = 1$ and $b ∉ (-1,1)$

$a = 2$ and $b ∉ \left(-\infty,1\right)$

$a = 5$ and $b ∈ (-\infty,1)$

Solution:

Point of intersection of two given lines is (1, 1). Since each of the two given lines contains a diameter of the given circle, therefore the point of intersection of the two given lines is the centre of the given circle.
Hence centre = (1,1)
$\therefore a^{2}-7a + 11= 1 \Rightarrow a = 2,5 \quad...\left(i\right)$
and $a^{2}-6a + 6=1 \Rightarrow a= 1,5\quad ...\left(ii\right)$
From both $\left(i\right)$ and $\left(ii\right), a = 5$
Now on replacing each of $\left(a^{2}- 7a +11\right)$ and $\left(a^{2} - 6a + 6\right)$ by 1, the equation of the given circle is $x^{2}+y^{2}-2x-2y+b^{3} +1 =0$
$\Rightarrow \left(x-1\right)^{2}+\left(y-1\right)^{2}+b^{3} = 1$
$\Rightarrow b^{3} = 1-\left[\left(x-1\right)^{2}+\left(y-1\right)^{2}\right]$
$\therefore b\in \left(-\infty, 1\right)$

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Q. If each of the lines $5x + 8y = 13$ and $4x-y = 3$ contains a diameter of the circle $x^2+y^2 - 2(a^2 - 7a +11)$ $x-2 (a^2-6a + 6)y + b^3 +1 = 0$, then :