Question

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

• A. $a=5$ and $b\in \left(-1,1\right)$
• B. $a=1$ and $b\in \left(-1,1\right)$
• C. $a=2$ and $b\in \left(-\infty ,1\right)$
• D. $a=5$ and $b\in \left(-\infty ,1\right)$

Answer 1

The correct option is D $a=5$ and $b\in \left(-\infty ,1\right)$

$x^2+y^{2}2(a^2-a+11)x-2(a^2-6a+6)y+6^3+1=0$

$\downarrow$

Centre $\left[\right(a^2-7a+11),(a^2-6a+6\left)\right]$

$\downarrow$

Should satisfy

$5x+8y=13$ & $4x-y=3$

$5(a$\downarrow$-7a+11)+8(a$\downarrow$-6a+6)-13=0$ &

$4(a^2-7a+11)-(a^2-6a+6)-3=0$

$13a^2-83a+90=0$ & $3a^2-22a+35=0$

radius $=\sqrt{(a^2-7a+11{)}^2+(a^2-6a+6{)}^2-(6^3+1)}$

$=\sqrt{(1{)}^2+(1{)}^2-(6^3+1)}$

$=\sqrt{1-6^3}$

$1-6^3>0$

$6^3<1$

$6\in (-\infty ,1)$

So,

$a=5$

$6\in (-\infty ,1)$