The length of the common chord of the two circles $x^{2} + y^{2} + 2 g x + c = 0$ and
$x^{2} + y^{2} + 2 f y - c = 0$ is

2 \sqrt{\frac{(g^{2} - c) (f^{2} - c)}{g^{2} + f^{2}}}

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

2 \sqrt{\frac{(g^{2} + c) (f^{2} + c)}{g^{2} + f^{2}}}

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

2 \sqrt{\frac{(g^{2} - c) (f^{2} + c)}{g^{2} + f^{2}}}

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

2 \sqrt{\frac{(g^{2} + c) (f^{2} - c)}{g^{2} + f^{2}}}

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

Open in App

Solution

The correct option is C $2 \sqrt{\frac{(g^{2} - c) (f^{2} + c)}{g^{2} + f^{2}}}$
$x^{2} + y^{2} + 2 g x + c = 0$
$C_{1} = c e n t r e (- g, 0), r_{1} = r a d i u s = \sqrt{g^{2} - c}$
and $x^{2} + y^{2} + 2 f y - c = 0$
center $\equiv (0, - f), r_{2} = r a d i u s = \sqrt{f^{2} + c}$
So, equation of common chord is
$2 g x - 2 f y + 2 c = 0$
Line $A B : g x - f y + c = 0$ ----(1)
$∴ C_{1} P = ∣ ∣ ∣ ∣ \frac{- g^{2} + c}{\sqrt{g^{2} + f^{2}}} ∣ ∣ ∣ ∣$
So, In $Δ C_{1} P A,$
$A P = \sqrt{g^{2} - c \frac{- (c - g^{2})}{g^{2} + f^{2}}}$
$= \sqrt{g^{2} - c} \sqrt{\frac{g^{2} + f^{2} - g^{2} + c}{g^{2} + f^{2}}}$
$A P = \sqrt{\frac{(g^{2} - c) (f^{2} + c)}{g^{2} + f^{2}}}$
So, length of chord $A B = 2 A P = 2 \sqrt{\frac{(g^{2} - c) (f^{2} + c)}{g^{2} + f^{2}}}$

Suggest Corrections

Similar questions

An equilateral triangle inscribed in the circle $x^{2} + y^{2} + 2 g x + 2 f y + c$ = 0 then its side is

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

(c > 0)

\frac{π}{2},

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

g^{2} < c < f^{2}

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

g^{2} + f^{2} = c,

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

Join BYJU'S Learning Program