Question

Four distinct points $(1,0),(0,1),(0,0)$ and $(t,t)$ are concyclic for

• A. $t=0$
• B. $t=1$
• C. $t=-1$
• D. None of these

Answer 1

Answer: (B)

$t=1$

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

Then as it passes through $(1,0)$

$\Rightarrow 1^2+0^2+2g+0+c=0$$\Rightarrow 1+2g+c=0$ ...(1)

As it passes through $(0,1)$

$\Rightarrow 0+1^2+0+2f+c=0$$\Rightarrow 1+2f+c=0$ ...(2)

And as it passes through $(0,0)$

$\Rightarrow 0+0+0+0+c=0\Rightarrow c=0$

Substituting this in (1) and (2) we get

$1+2g=0\Rightarrow g=\frac{-1}{2}$ and $1+2f=0\Rightarrow f=\frac{-1}{2}$

$\therefore$ Equation of circle is

$x^2+y^2+2\left(\frac{-1}{2}\right)x+2\left(\frac{-1}{2}\right)y+0=0$

$\Rightarrow x^2+y^2-x-y=0$

It is clear that it will passes through $(1,1)$

$\therefore t=1.$