Discussion Forum

The circle passing through the point (- 1, 0)and touching the Y-axis at (0 2), also passes through the point

Answer:

Explanation:

Equation of circle passing ihrough a point (x₁,y₁) and touching the straight
line L. is given by

 $(x-x_{1})^{2}+(y-y_{1})^{2}= \lambda L=0$

$\therefore$   Equation sf circle passing through (0, 2) and touching x = 0.

Now,  $(x-0)^{2}+(y-2)^{2}+\lambda x=0$...(i)

 Also, it passes through (-1,0)

 so, $1+4-\lambda=0 \Rightarrow  0 \Rightarrow \lambda=5$

 Eq.(i) becomes 

       $x^{2}+y^{2}-4y+4+5x=0$

 $\Rightarrow$  $x^{2}+y^{2}+5x-4y+4=0$

 For x- intercept put y=0

  $x^{2}+5x+4=0$

$\Rightarrow$ $(x+1)(x+4)=0$ 

$\therefore$   x=-1,-4