MyQuestionIcon

1

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

Touching Circles Theorem

Two circles ...

Question

Two circles $x^{2} + y^{2} - 2 k x = 0$ and $x^{2} + y^{2} - 4 x - 8 y + 16 = 0$ touch each other externally. Then, $k$ is

A

4

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

- 4

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

Open in App

Solution

The correct option is C $2$
General equation of circle is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
Centre $(- g, - f)$ Radius $= \sqrt{g^{2} + f^{2} - c}$
$x^{2} + y^{2} - 2 k x = 0$ $x^{2} + y^{2} - 4 x - 8 y + 16 = 0$
$C_{1} = (k, 0) C_{2} = (2, 4)$
$R_{1} = \sqrt{{(- k)}^{2}} = k R_{2} = \sqrt{{(- 2)}^{2} + {(- 4)}^{2} - 16} = 2$
If both circles touch each other externally then $C_{1} C_{2} = R_{1} + R_{2}$
$\sqrt{{(k - 2)}^{2} + {(0 - 4)}^{2}} = k + 2$
${(k - 2)}^{2} + 16 = {(k + 2)}^{2}$
$k^{2} + 4 - 4 k + 16 = k^{2} + 4 + 4 k$
$8 k = 16 \Rightarrow k = 2$

Suggest Corrections

thumbs-up

0

Similar questions

Q. Examine if the two circles

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

x^{2} + y^{2} - 8 y - 4 = 0

touch each other externally or internally

Q. If two circles

x^{2} + y^{2} - 2 a x + c^{2} = 0

x^{2} + y^{2} - 2 b y + c^{2} = 0

touch each other externally, then

Q. If the circles

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

x^{2} + y^{2} - 8 y - k = 0

touches each other internally, then the possible value of

k

The two circles $x^{2} + y^{2} - 4 y = 0$ and $x^{2} + y^{2} - 8 y = 0$

Q. lf the circles

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

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

touch internally then

K =

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Touching Circles Theorem

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Touching Circles Theorem

Standard X Mathematics

Join BYJU'S Learning Program

CrossIcon