English

The angle between two circles x2+y2-12x-6y+41 = 0 and x2+y2+kx+6y-59 = 0 is 45. Find the value of K: -

Advertisements
Advertisements

Question

The angle between two circles `x^2 + y^2 - 12x - 6y + 41` = 0 and `x^2 + y^2 + kx + 6y - 59` = 0 is 45. Find the value of K:

Options

  • ± 3

  • ± 4

  • 4

  • – 4

MCQ

Solution

± 4

Explanation:

Let θ is the angle between the circles.

First Circle: `x^2 + y^2 - 12x - 6y + 41` = 0

∴ `g_1 = (-6), f_1 = (-3), c_1 = 41`

Second Circle: `x^2 + y^2 + kx + 6y - 59` = 0

∴  `g_2 = k/2; f_2 = 3, c_2 = - 59`

Now, `cos theta = (c_1 + c_2 - 2g_1*g_2 - 2f_1*f_2)/(2r_1r_2)`

⇒ `cos 45^circ = (41 - 59 - 2(-6) k/2 - 2(-3)(3))/(2sqrt(36 + 9 - 41) xx sqrt(k^2/4 + 9 + 59)`

⇒ `1/sqrt(2) = (- 18 + 6k + 18)/(2.2 sqrt(k^2/4 + 68)`

⇒ `1/sqrt(2) = (6k)/(4sqrt(k^2/4 + 68)`

Squaring and cross multiplying :

`4[k^2/4 + 68] = 18k^2` 

⇒ `(2[k^2 + 272])/4 = 9k^2`

⇒ `k^2 + 272 = 18k^2`

⇒ `17k^2` = 272

⇒ `k^2 = 272/17` = 16

⇒ `k` = ± 4

shaalaa.com
Geometry (Entrance Exam)
  Is there an error in this question or solution?
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×