Question

The distinct points $A_1(0,0),A_2(0,1),A_3(1,0)$ and $A_4(2a,3a)$ are concylic then

• A. $a<0$
• B. $aϵ(0,1)$
• C. $a$ can attain only rational values
• D. $Noneofthese$

Answer 1

The correct option is A $a<0$

We have,

$A_1(0,0),A_2(0,1),A_3(1,0),A_4(2a,3a)$

Now,

We know that,

The equation of circle be $x^2+y^2+2gx+2fy+c=0$

At the point $A_1(0,0)$

So,

$c=0$

At the point $A_2(0,1)$

So,

$0+1+0+2f+c=0$

$\Rightarrow 1+2f+c=0$

$Ifc=0$

$So,$

$\Rightarrow 2f+1=0$

$\Rightarrow f=\frac{-1}{2}$

At the point $A_3(1,0)$

So,

$1+0+2g+0+0=0$

$\Rightarrow g=\frac{-1}{2}$

The equation of circle is

$x^2+y^2+2gx+2fy+c=0$

Lies the point $(2a,3a)$

So,

$4a^2+9a^2+2\times \frac{-1}{2}\times 2a+2\times \frac{-1}{2}\times 3a+0=0$

$\Rightarrow 13a^2-2a-3a=0$

$\Rightarrow 13a^2-5a=0$

$\Rightarrow a(13a-5)=0$

$\Rightarrow a=0,13a-5=0$

$\Rightarrow a=0,\frac{5}{13}$

Hence, this is the answer.