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

Orthogonality of Two Circles

The circle ...

Question

The circle $x^{2} + y^{2} + 4 x + 6 y + 3 = 0$ and $2 (x^{2} + y^{2}) + 6 x + 4 y + c = 0$ will cut orthogonally of c $= ?$

4

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

18

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

12

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

16

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

Open in App

Solution

The correct option is B $18$
$x^{2} + y^{2} + 4 x + 6 y + 3 = 0 g_{1} = 2, f_{1} = 3, c_{1} = 3 a n d \Rightarrow 2 (x^{2} + y^{2}) + 6 x + 4 y + c = 0 f_{1} = \Rightarrow x^{2} + y^{2} + 3 x + 2 y + \frac{c}{2} = 0 \Rightarrow g_{2} = \frac{3}{2}, f_{2} = 1, c_{2} = \frac{c}{2} i f t w o c i r c l e s c u t s o r t h o g o n a l l y t h e n \Rightarrow 2 f_{1} f_{2} + 2 g_{1} g_{2} = c_{1} + c_{2} \Rightarrow 2 \times 3 \times 1 + 2 \times 2 \times \frac{3}{2} = 3 + \frac{c}{2} \Rightarrow 6 + 6 = 3 + \frac{c}{2} \Rightarrow 12 - 3 = \frac{c}{2} \Rightarrow 9 = \frac{c}{2} ∴ c = 18$

Suggest Corrections

Similar questions

x^{2} + y^{2} + 4 x + 6 y + 3 = 0

2 (x^{2} + y^{2}) + 6 x + 4 y + c = 0

c

x^{2} + y^{2} - 4 x - 6 y - 12 = 0

x^{2} + y^{2} + 6 x + 4 y - 12 = 0

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

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

x^{2} + y^{2} - 6 x + 4 y - 12 = 0

S

x^{2} + y^{2} - 3 x - 6 y + 14 = 0, x^{2} + y^{2} - x - 4 y + 8 = 0

x^{2} + y^{2} + 2 x - 6 y + 9 = 0

Join BYJU'S Learning Program