CameraIcon

SearchIcon

MyQuestionIcon

1

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

Number of Common Tangents to Two Circles in Different Conditions

The number of...

Question

The number of common tangents to the following pairs of circles $x^{2} + y^{2} + 4 x - 6 y - 3 = 0$ and $x^{2} + y^{2} + 4 x - 2 y + 4 = 0$ is

A

0

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

B

1

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

C

2

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

D

3

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

Open in App

Solution

The correct option is A $0$
Given circles are
$x^{2} + y^{2} + 4 x - 6 y - 3 = 0$ and $x^{2} + y^{2} + 4 x - 2 y + 4 = 0$
The centre and radius are,
$C_{1} (- 2, 3), r_{1} = 4 C_{2} (- 2, 1), r_{2} = 1 C_{1} C_{2} = 2 \Rightarrow C_{1} C_{2} < r_{1} - r_{2}$

Therefore, one circle lies completely inside the other.
Hence, the number of common tangents is $0.$

Suggest Corrections

thumbs-up

0

Similar questions

Q. The number of common tangents that can be drawn to the circles

x^{2} + y^{2} - 4 x - 6 y - 3 = 0

x^{2} + y^{2} = 2 x + 2 y + 1 = 0

Q. Find the transverse common tangents of the circles:

x^{2} + y^{2} - 4 x - 10 y + 28 = 0

x^{2} + y^{2} + 4 x - 6 y + 4 = 0

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

Q. Find the transverse common tangents of the circle

x^{2} + y^{2} - 4 x - 10 y + 28 = 0

x^{2} + y^{2} + 4 x - 6 y + 4 = 0

Q. Find the equation to the pair of transverse common tangents to the circles

x^{2} + y^{2} - 4 x - 10 y + 28 = 0 a n d x^{2} + y^{2} + 4 x - 6 y + 4 = 0

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Number of Common Tangents to Two Circles in Different Conditions

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon