Circle 4 Question 10

10. 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: 10. (d)

Solution:

  1. Let $C _1(h, k)$ be the centre of the required circle. Then,

$$ \begin{aligned} & & \sqrt{(h-0)^{2}+(k-0)^{2}} & =\sqrt{(h-1)^{2}+(k-0)^{2}} \\ \Rightarrow & & h^{2}+k^{2} & =h^{2}-2 h+1+k^{2} \\ \Rightarrow & & -2 h+1 & =0 \Rightarrow h=1 / 2 \end{aligned} $$

Since, $(0,0)$ and $(1,0)$ lie inside the circle $x^{2}+y^{2}=9$. Therefore, the required circle can touch the given circle internally.

i.e.

$$ C _1 \cdot C _2=r _1 \sim r _2 $$

$$ \Rightarrow \quad \sqrt{h^{2}+k^{2}}=3-\sqrt{h^{2}+k^{2}} $$

$$ \begin{aligned} & \Rightarrow \quad 2 \sqrt{h^{2}+k^{2}}=3 \quad \Rightarrow 2 \sqrt{\frac{1}{4}+k^{2}}=3 \\ & \Rightarrow \quad \sqrt{\frac{1}{4}+k^{2}}=\frac{3}{2} \Rightarrow \frac{1}{4}+k^{2}=\frac{9}{4} \end{aligned} $$

$$ \Rightarrow \quad k^{2}=2 \Rightarrow k= \pm \sqrt{2} $$



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