Question

If the circle passing through the points $(0,0),(0,1)and(1,0)$ also passes through the point $(\lambda ,\frac{1}{\lambda })$ then $\lambda$ is

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

The correct option is A $1$

The general form of a circle is $x^2+y^2+2gx+2fy+c=0$

Now, if it passes through $(0,0)$, then $0^2+0^2+2g\left(0\right)+2f\left(0\right)+c=0⟹c=0$

Now, if it passes through $(0,1)$, then $0^2+1^2+2g\left(0\right)+2f\left(1\right)=0⟹f=-\frac{1}{2}$

Now, if it passes through $(1,0)$, then $1^2+0^2+2g\left(1\right)+2f\left(0\right)=0⟹g=-\frac{1}{2}$

Hence, the equation of the circle is $x^2+y^2-x-y=0$.

Also, this circle should pass through $(\lambda ,\frac{1}{\lambda })$, so

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

$\therefore {\lambda }^4+1-{\lambda }^3-\lambda =0$

$\therefore \lambda =1$ (only real solution)

This is the required solution.