SATHEE: Circle 4 Question 10

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) $(1 / 2, 3 / 2)$

(d) $(1 / 2, - 2^{1 / 2})$

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Answer:

Correct Answer: 10. (d)

Solution:

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.

$C_{1} \cdot C_{2} = r_{1} \sim r_{2}$

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

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

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

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

Answer:

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Circle 4 Question 10

10. The centre of a circle passing through the points (0,0), (1,0) and touching the circle x2+y2=9 is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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10. The centre of a circle passing through the points $(0, 0)$ , $(1, 0)$ and touching the circle $x^{2} + y^{2} = 9$ is