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

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Solution:

## Equation of circle passing through a point $(x_1, y_1)$ and touching the straight line L, is given by $( x - x_1 )^2 + ( y - y_1)^2 + \lambdaL = 0$ $\therefore$ Equation of circle passing through (0, 2) and touching x=0 $\Rightarrow \, \, \, \, \, \, \, ( x -0)^2+ (y -2 )^2+ \lambdax = 0\hspace15mm ...(i)$ Also, it passes through $(-1 ,0 ) \Rightarrow 1 + 4 - \lambda = 0$ $\therefore$ \lambda=5 Eq. (i) becomes, $\hspace22mm x^2 + y^2 - 4y + 4 + 5x = 0$ $\Rightarrow\hspace15mm x^2 + y^2 - 5x - 4y+4 =0$, For x-intercept put $y = 0 \Rightarrow x^2 + 5x + 4 = 0,$ $(x + 1) (x+4)=0$ $\therefore$ $\hspace20mm$x=-1, -4 Hence, (d) option (-4, 0).

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