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If each of the lines $5 x+8 y=13$ and $4 x-y=3$ contains a diameter of the circle $x^2+y^2-2\left(a^2-7 a+11\right)$
$x-2\left(a^2-6 a+6\right) y+b^3+1=0$, then :
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$x-2\left(a^2-6 a+6\right) y+b^3+1=0$, then :
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Verified Answer
The correct answer is:
$a=5$ and $b \in(-\infty, 1)$
$a=5$ and $b \in(-\infty, 1)$
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 $=(1,1)$
$$
\therefore a^2-7 a+11=1 \Rightarrow a=2,5
$$
and $a^2-6 a+6=1 \Rightarrow a=1,5$
From both (i) and (ii), $a=5$
Now on replacing each of $\left(a^2-7 a+11\right)$ and $\left(a^2-6 a+6\right)$ by 1 , the equation of the given circle is $x^2+y^2-2 x-2 y+b^3+1=0$
$$
\begin{aligned}
& \Rightarrow(x-1)^2+(y-1)^2+b^3=1 \\
& \Rightarrow b^3=1-\left[(x-1)^2+(y-1)^2\right] \\
& \therefore b \in(-\infty, 1)
\end{aligned}
$$
Hence centre $=(1,1)$
$$
\therefore a^2-7 a+11=1 \Rightarrow a=2,5
$$
and $a^2-6 a+6=1 \Rightarrow a=1,5$
From both (i) and (ii), $a=5$
Now on replacing each of $\left(a^2-7 a+11\right)$ and $\left(a^2-6 a+6\right)$ by 1 , the equation of the given circle is $x^2+y^2-2 x-2 y+b^3+1=0$
$$
\begin{aligned}
& \Rightarrow(x-1)^2+(y-1)^2+b^3=1 \\
& \Rightarrow b^3=1-\left[(x-1)^2+(y-1)^2\right] \\
& \therefore b \in(-\infty, 1)
\end{aligned}
$$
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