Question

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

• A.
• B.
• C.
• D. (-4, 0)

Answer 1

The correct option is D

(-4, 0)

Circle touching y-axis at (0, 2) equation of circle is

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

This passes through (-1, 0)

So $(-1,-0{)}^2+(0-2{)}^2+\lambda (-1)=0$

$1+4-\lambda =0$

$\lambda =5$

Substituting the value of λin equation (1)

$x^2+(y-2{)}^2+5x=0$

$x^2+y^2+5x-4y+4=0$- - - - - - (2)

We have to check it from the given options.

We have $y=0,2,\frac{5}{2}$

let's take y = 0

Then, from equation (2)

$x^2+0+5x-0+4=0$

$(x+1)(x+4)=0$

$x=-1,-4$

Circle passes through points $(-1,0)(-4,0)$

We see only (-4, 0) is there in the options. So, circle passes through (-4, 0).

Option D is correct.

Similarly, we can substitute $y=2,\frac{1}{2}$ in equation (2) and find the value of x. we find that none of other points are given in the options. Only option D is correct.