Question

Circle $x^2+y^2-2x+\lambda x-1=0$ passes through two fixed points, then coordinates of the points are

• A. $(0,\pm 1)$
• B. $(\pm 1,0)$
• C. $(0,1)$ and $(0,2)$
• D. $(0,-1)$ and $(0,-2)$

Answer 1

The correct option is A $(0,\pm 1)$

Solution :- given

$\Rightarrow x^2+y^2-2x+\lambda x-1=0$

$\Rightarrow x^2-x(2-\lambda )+y^2=1$

$\Rightarrow x^2-2.x.\frac{(2-\lambda )}{2}+\frac{(2-\lambda {)}^2}{4}-\frac{(2-\lambda {)}^2}{4}+y^2=1$

$\Rightarrow {(x-\frac{(2-\lambda )}{2})}^2+y^2=1+\frac{(2-\lambda {)}^2}{4}$

$\Rightarrow$ as circle passes through 2 fixed points

These points will be given when $x=0$

Now putting $x=0$

$\Rightarrow {(-\frac{(2-\lambda )}{2})}^2+y^2=1+\frac{(2-\lambda {)}^2}{4}$

$\Rightarrow y^2+\frac{(2-\lambda {)}^2}{4}=1+\frac{(2-\lambda {)}^2}{4}$

$\Rightarrow y=\pm 1$

$\therefore$ The reqd points is $=(0,\pm 1)$ (Ans)