The two circles $x^{2} + y^{2} = a x a n d x^{2} + y^{2} = c^{2} (c > 0)$ touch each other if

2 | a | = c

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

| a | = c

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

a = 2 c

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

| a | = 2 c

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

Open in App

Solution

The correct option is A $2 | a | = c$
If two circles touch each other

$⟹ r_{1} + r_{2} = c_{1} c_{2}$ (externally)

Or

$⟹ | r_{1} - r_{2} | = c_{1} c_{2}$ (internally)

$⟹ a + c = a$

But $c \neq 0$

$| a - c | = a$

$⟹ c \neq 0, c = 2 | a |$

Correct answer is $c \neq 0, c = 2 | a |$

Suggest Corrections

Similar questions

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

x^{2} + y^{2} = c^{2} (c > 0)

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

x^{2} + y^{2} = c^{2} (c > 0)

x^{2} + y^{2} = c^{2} (c > 0)

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

x^{2} + y^{2} = c^{2} (c > 0)

\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c}

\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c^{2}}

\frac{1}{a} + \frac{1}{b} = \frac{2}{c}

Join BYJU'S Learning Program