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

Orthogonality of Two Circles

Find the valu...

Question

Find the value of k, if the circles $x^{2} + y^{2} + 4 x + 8 = 0$ and $x^{2} + y^{2} - 16 x + k = 0$ are orthogonal.

Open in App

Solution

Fact:
Two circles are orthogonal if $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$

Here $g_{1} = 2, f_{1} = 0, c_{1} = 8, g_{2} = - 8, f_{2} = 0, c_{2} = k$
$∴ 2 (2) (- 8) + 2 (0) (0) = 8 + k$
$\Rightarrow - 32 = k + 8$
$\Rightarrow k = - 32 - 8 = - 40$

Suggest Corrections

Similar questions

x^{2} + y^{2} - 5 x - 14 y - 34 = 0, x^{2} + y^{2} + 2 x + 4 y + k = 0

^{'} k^{'} .

k

x^{2} + y^{2} + 5 x + 3 y + 7 = 0

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

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

k

2 x^{2} + 2 y^{2} - 3 x + 6 y + k = 0

x^{2} + y^{2} - 4 x + 10 y + 16 = 0

k

Join BYJU'S Learning Program