Circle Ques 73
- The centre of a circle passing through the points $(0,0)$, $(1,0)$ and touching the circle $x^{2}+y^{2}=9$ is
$(1992,2 M)$
(a) $(3 / 2,1 / 2)$
(b) $(1 / 2,3 / 2)$
(c) $(1 / 2,1 / 2)$
(d) $\left(1 / 2,-2^{1 / 2}\right)$
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Answer:
Correct Answer: 73.(d)
Solution:
Formula:
- Let $C _1(h, k)$ be the centre of the required circle. Then,
$$ \begin{alignedat} & & \sqrt{(h-0)^{2}+(k-0)^{2}} & =\sqrt{(h-1)^{2}+(k-0)^{2}} \\ \Rightarrow & & h^{2}+k^{2} & =h^{2}-2 h+1+k^{2} \\ \Rightarrow & & -2 h+1 & =0 \Rightarrow h=1 / 2 \end{aligned} $$
Since, $(0,0)$ and $(1,0)$ lie inside the circle $x^{2}+y^{2}=9$. Therefore, the required circle can touch the given circle externally.
i.e.
$$ C _1 \cdot C _2=r _1 \cdot r _2 $$
$$ \Rightarrow \quad \sqrt{h^{2}+k^{2}}=3-\sqrt{h^{2}+k^{2}} $$
$$ \begin{alignedat} & \Rightarrow \quad 2 \sqrt{h^{2}+k^{2}}=3 \quad \Rightarrow 2 \sqrt{\left(\frac{1}{2}\right)^{2}+k^{2}}=3 \\ & \Rightarrow \quad \sqrt{\frac{1}{4}+k^{2}}=\frac{3}{2} \Rightarrow \frac{1}{4}+k^{2}=\frac{9}{4} \end{aligned} $$
$$ \Rightarrow \quad k^{2}=2 \Rightarrow k= \pm \sqrt{2} $$
BETA
