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

Options

A. `(-(3)/(2),0)`
B. `(-5/2,2)`
C. `(-(2)/(2),(5)/(2))`
D. (-4,0)

Answer

D

Explanation

Equation of circle passing through a point (`(x_1,y_1)` and touching straight line L, is given by `(x-x_1)^(2)+(y-y_1)^(2)+lambdal=0`
equation of circle passing through (0,2) and tounching x=0 is `(x-0)^(2)+(y-2)^(2)lambdal=0` also, it passes through (-1,0).
`rArr1+4-lambda=0` :. lambda=5
Eq.(i) becomes, `x^(2)+y^(2)-4y+4+5x=0`
`x^(2)+y^(2)+5x+4y+4=0` for x-intercept, put y=0.
`x^(2)+5x+4=0`
`(x+1)(x+4)=0`
:. x=-1,-4 Hence, required point is (-4,0) or,(-1,0).