CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

COM of Center of Masses

C1:x2+y2+2x-3...

Question

$C_{1} : x^{2} + y^{2} + 2 x - 3 = 0, C_{2} : x^{2} + y^{2} - 2 x - 3 = 0, C_{3} : x^{2} + y^{2} - 2 y - 3 = 0,$ are three circles

A

C_{1}

passes through the centre of

C_{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

B

C_{2}

passes through the centre of

C_{1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C

centres of the three circles form an equilateral triangle

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

sum of the radii of

C_{1}

and

C_{2}

is twice the radius of

C_{3}

.

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct options are
A $C_{1}$ passes through the centre of $C_{2}$
C $C_{2}$ passes through the centre of $C_{1}$
D sum of the radii of $C_{1}$ and $C_{2}$ is twice the radius of $C_{3}$ .
Centers of $C_{1}$ , $C_{2}$ and $C_{3}$ are $(- 1, 0)$ , $(1, 0)$ and $(0, 1)$ respectively.
Since, $(1, 0)$ satisfies $C_{1}$
Therefore, $C_{1}$ passes through the centre of $C_{2}$
and $(- 1, 0)$ satisfies $C_{2}$
Therefore, $C_{2}$ passes through the centre of $C_{1}$
now, $r_{1} = \sqrt{1 + 3} = 2$
Similarly, $r_{2} = r_{3} = 2$
Therefore, $r_{1} + r_{2} = 2 r_{3}$

Suggest Corrections

thumbs-up

0

Similar questions

C_{1} : x^{2} + y^{2} = 1; C_{2} : (x - 10)^{2} + y^{2} = 1

C_{3} : x^{2} + y^{2} - 10 x - 42 y + 457 = 0

be three circles. A circle C has been drawn to touch circles

C_{1}

C_{2}

C_{3}

internally. Now circles

C_{1}, C_{2}

C_{3}

start rolling on the circumference of circle C in anticlockwise direction with constant speed. The centroid of the triangle formed by joining the centres of rolling circles

C_{1}, C_{2}

C_{3}

Let $c_{1} : x^{2} + y^{2} = 1; C_{2} : (x - 10)^{2} + y^{2} = 1 a n d C_{3}; x^{2} + y^{2} - 10 x - 42 y + 457 = 0$ be three circle.A circle C has been drawn to touch circles $C_{1}$ and $C_{2}$ externally and $C_{3}$ internally. Now circles $C_{1}, C_{2}$ and $C_{3}$ start rolling on the circumference of circle C in anticlockwise direction with constant speed. The centroid of the triangle formed by joining the centres of rolling circles $C_{1}, C_{2}$ and $C_{3}$ lies on

Q. Assertion :Consider circles

C_{1} \equiv x^{2} + y^{2} - 6 x - 4 y + 9 = 0

C_{2} \equiv x^{2} + y^{2} - 8 x - 6 y + 23 = 0

C_{1}

bisects the circumference of the circle

C_{2}

Reason: Centre of

C_{1}

C_{2}

Q. For the circles

C_{1} : x^{2} + y^{2} + 2 x + c = 0

C_{2} : x^{2} + y^{2} + 2 y + c = 0

which of the following is/are correct?

c_{1} : x^{2} + y^{2} - 2 x - 2 y + 3 = 0

c_{2} : x^{2} - y^{2} - 2 x - 2 y + 7 = 0

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Integrating Solids into the Picture

PHYSICS

Watch in App

explore_more_icon

Explore more

COM of Center of Masses

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon