JEE Advance - Mathematics (2011 - Paper 2 Offline - No. 4)

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

$$\left( { - {3 \over 0},0} \right)$$

$$\left( { - {5 \over 2},2} \right)$$

$$\left( { - {3 \over 0},\,{5 \over 2}} \right)$$

(- 4, 0)

Circle touching y-axis at (0, 2) is $${(x - 0)^2} + {(y - 2)^2} + \lambda x = 0$$ passes through ($$-$$1, 0).

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

Hence, $${x^2} + {y^2} + 5x - 4y + 4 = 0$$

Substituting $$y = 0 \Rightarrow x = - 1,\, - 4$$.

Hence, the Circle passes through ($$-$$4, 0).