Circle 4 Question 6
6. The circle passing through the point $(-1,0)$ and touching the $Y$-axis at $(0,2)$ also passes through the point
(2011)
(a) $-\frac{3}{2}, 0$
(b) $-\frac{5}{2}, 2$
(c) $-\frac{3}{2}, \frac{5}{2}$
(d) $(-4,0)$
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Answer:
Correct Answer: 6. (d)
Solution:
- Equation of circle passing through a point $\left(x _1, y _1\right)$ and touching the straight line $L$, is given by
$$ \left(x-x _1\right)^{2}+\left(y-y _1\right)^{2}+\lambda L=0 $$
$\therefore$ Equation of circle passing through $(0,2)$ and touching $x=0$
$\Rightarrow \quad(x-0)^{2}+(y-2)^{2}+\lambda x=0$
Also, it passes through $(-1,0) \Rightarrow 1+4-\lambda=0$
$\therefore \lambda=5$
Eq. (i) becomes,
$$ \begin{aligned} x^{2}+y^{2}-4 y+4+5 x & =0 \\ \Rightarrow \quad x^{2}+y^{2}+5 x-4 y+4 & =0 \end{aligned} $$
For $x$-intercept put $y=0 \Rightarrow x^{2}+5 x+4=0$,
$$ \begin{aligned} & & (x+1)(x+4) & =0 \\ & \therefore & x & =-1,-4 \end{aligned} $$
Hence, (d) option $(-4,0)$.
