MyQuestionIcon

1

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

Real Valued Functions

The number of...

Question

The number of distinct points common to the graphs of $x^{2} + y^{2} = 4$ and $y^{2} = 4$ is

A

0

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

B

1

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

C

2

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

D

3

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

E

4

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

Open in App

Solution

The correct option is D 2
The graphs are $x^{2} + + y^{2} = 4$ and $y^{2} = 4$
$\Rightarrow x^{2} + 4 = 4$
$\Rightarrow x^{2} = 0$
$\Rightarrow x = 0$
Therefore $y = \pm 2$
Therefore, the points common to both are $(0, 2), (0, - 2)$ .
The number of points is $2$ .

Suggest Corrections

thumbs-up

0

Similar questions

Q. The number of common tangents to the circles

x^{2} + y^{2} = 4

x^{2} + y^{2} - 4 x + 2 y - 4 = 0

Q. Find the co - ordinates of the circles

x^{2} + y^{2} - 4 x - 2 y = 4

x^{2} + y^{2} - 12 x - 8 y = 12

touch each other. Also find equations of common tangents touching the circles in distinct points.

Q. The common tangent to the circles

x^{2} + y^{2} = 4

x^{2} + y^{2} + 6 x + 8 y - 24 = 0

also passes through the point:

Q. The common tangent to the circles

x^{2} + y^{2} = 4

x^{2} + y^{2} + 6 x + 8 y - 24 = 0

also passes through the point :

Q. A transverse common tangent to the circles

x^{2} + y^{2} + 4 x + 2 y = 4

x^{2} + y^{2} - 4 x - 2 y + 4 = 0

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Real Valued Functions

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Real Valued Functions

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon