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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
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Geometry (Entrance Exam)
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