Question

If the circle $x^2+y^2-2(a^2-3a+1)x-2(a^2-5a+3)y+b^2+1=0$ has diameters 3x - 4y - 1 = 0 and 8x - 3y + 5 = 0, then

• A. $a=1,bϵR(-\infty ,\infty )$
• B. $a=2,bϵ[-\sqrt{3},\sqrt{3}]$
• C. $a=1,bϵ[-\sqrt{3},\sqrt{3}]$
• D. $a=1,bϵ(-1,1)$

Answer 1

The correct option is D $a=1,bϵ(-1,1)$

Equation of diameters are 3x - 4y - 1 = 0
8x - 3y + 5 = 0,
$\therefore$ solving, we get centre = (-1, -1)
But centre of circle $x^2+y^2-2(a^2-3a+1)x-2(a^2-5a+3)y+b^2+1=0$ is $(a^2-3a+1,a^2-5a+3)=(-1,-1)$
$\Rightarrow a^2-3a+2=0and{a}^2-5a+4=0$
$\Rightarrow a=1$
Radius = $\sqrt{1+1-(b^2+1)}=\sqrt{1-b^2}$ is defined when $1-b^2$ > 0 $\Rightarrow$ $bϵ(-1,1)$