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

Number of Common Tangents to Two Circles in Different Conditions

lf the circle...

Question

lf the circles $x^{2} + y^{2} + 2 x + c = 0$ and $x^{2} + y^{2} + 2 y + c = 0$ touch each other then $c =$

\frac{1}{2}

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

\frac{1}{4}

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

2

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

4

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

Open in App

Solution

The correct option is D $\frac{1}{2}$
From point (1),
$S_{1} : x^{2} + y^{2} + 2 x + c = 0 \Rightarrow C_{1} \equiv (- 1, 0)$ & $r_{1} = \sqrt{1 - c}$
$S_{1} : x^{2} + y^{2} + 2 y + c = 0 \Rightarrow C_{2} \equiv (0, - 1)$ & $r_{2} = \sqrt{1 - c}$
case - I
Circles touch each other externally,
$C_{1} C_{2} = r_{1} + r_{2}$
$\Rightarrow \sqrt{1 + 1} = 2 \sqrt{1 - c}$
$2 = 4 - 4 c$
$4 c = 2$
$c = \frac{1}{2}$
case - II
circle touch each other internally.
$C_{1} C_{2} = r_{1} + r_{2}$
$\sqrt{2} = C_{1} C_{2} = 0$
X not possible.

Suggest Corrections

Similar questions

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

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

\frac{1}{c} =

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

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

If the circles $x^{2} + y^{2} + 2 a x + c = 0 a n d x^{2} + y^{2} + 2 b y + c = 0$ touch each other, then c is _________.where $a & b ϵ R$ .

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

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

I

x^{2} + y^{2} - 6 x - 8 y + 12 = 0, a) 2

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

k =

I I

x^{2} + y^{2} - 2 x + 3 y + k = 0, b) - 10

x^{2} + y^{2} + 8 x - 6 y - 7 = 0

k =

I I I

x^{2} + y^{2} + 2 x - 2 y + 4 = 0, c) - 24

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

k =

Join BYJU'S Learning Program