Length of the tangent drawn from any point on the circle $x^{2} + y^{2} + 2 g x + 2 f y + c_{1} = 0$ to the circle is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$

\sqrt{c_{1} - c}

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

\sqrt{c - c_{1}}

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

\sqrt{c_{1} + c}

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

None of these

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

Open in App

Solution

The correct option is B $\sqrt{c - c_{1}}$
Suppose $(x_{1}, y_{1})$ be any point on first circle from which tangent is to be drawn, then
$x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c_{1} = 0$ ….(i)
and also length of tangent
$= \sqrt{S_{2}} = \sqrt{x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c}$ ….(ii)
From (i), we get (ii) as $\sqrt{c - c_{1}} .$

Suggest Corrections

Similar questions

x^{2} + y^{2} + 2 g x + 2 f y + c_{1} = 0

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

x^{2} + y^{2} + 2 g x + 2 f y + d = 0

x^{2} + y^{2} + 2 g x + 2 f y + p = 0

x^{2} + y^{2} + 2 g x + 2 f y + q = 0

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

x^{2} + y^{2} + 2 g x + 2 f y + c s i n^{2} α + (g^{2} + f^{2}) c o s^{2} α = 0

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

x^{2} + y^{2} + 2 g x + 2 f y + c {sin}^{2} α + (g^{2} + f^{2}) {cos}^{2} α = 0

2 α

Join BYJU'S Learning Program