Question

please solve the given question ​

Answer 1

Answer:

(a),(d)

Step-by-step explanation:

$\textbf{\underline{\underline{Note:}}}$

Two circles are said to intersect orthogonally if their angle of intersection is a right angle.

Let the equations of the two circles be:

x² + y² + 2g₁x + 2f₁y + c₁ = 0 and x² + y² + 2g₂x + 2f₂y + c₂ = 0.

The two circles will intersect orthogonally if 2g₁g₂ + 2f₁f₂ = c₁ + c₂.

Now,

Given circles are:

x² + y² + 2kx + 2y + 6 = 0

x² + y² + 2kx + k = 0.

They intersect orthogonally.

2g₁g₂ + 2f₁f₂ = c₁ + c₂

2(g₁g₂ + f₁f₂) = c₁+c₂

⇒ 2(-k * -k + 0) = 6 + k

⇒ 2(k² + 0) = 6 + k

⇒ 2k² = 6 + k

⇒ -2k² + k + 6 = 0

⇒ 2k² - k - 6 = 0

⇒ 2k² + 3k - 4k - 6 = 0

⇒ k(2k + 3) - 2(2k + 3) = 0

⇒ (k - 2)(2k + 3) = 0

⇒ k = 2, -3/2

Hope it helps!

Answer 2

Hey Dear !

Answer :- a and d .

=) Let the equations of the two circles be:

the equations of the two circles be:x² + y² + 2g₁x + 2f₁y + c₁ = 0 and x² + y² + 2g₂x + 2f₂y + c₂ = 0.

the equations of the two circles be:x² + y² + 2g₁x + 2f₁y + c₁ = 0 and x² + y² + 2g₂x + 2f₂y + c₂ = 0.The two circles will intersect foursquare if 2g₁g₂ + 2f₁f₂ = c₁ + c₂.

if 2g₁g₂ + 2f₁f₂ = c₁ + c₂.Now ,

Given circles here :-

x² + y² + 2kx + 2y + 6 = 0

x² + y² + 2kx + 2y + 6 = 0x² + y² + 2kx + k = 0.

x² + y² + 2kx + 2y + 6 = 0x² + y² + 2kx + k = 0.They intersect foursquare .

2g₁g₂ + 2f₁f₂ = c₁ + c₂

2g₁g₂ + 2f₁f₂ = c₁ + c₂2(g₁g₂ + f₁f₂) = c₁+c₂

2g₁g₂ + 2f₁f₂ = c₁ + c₂2(g₁g₂ + f₁f₂) = c₁+c₂⇒ 2(-k * -k + 0) = 6 + k

2g₁g₂ + 2f₁f₂ = c₁ + c₂2(g₁g₂ + f₁f₂) = c₁+c₂⇒ 2(-k * -k + 0) = 6 + k⇒ 2(k² + 0) = 6 + k

2g₁g₂ + 2f₁f₂ = c₁ + c₂2(g₁g₂ + f₁f₂) = c₁+c₂⇒ 2(-k * -k + 0) = 6 + k⇒ 2(k² + 0) = 6 + k⇒ 2k² = 6 + k

2g₁g₂ + 2f₁f₂ = c₁ + c₂2(g₁g₂ + f₁f₂) = c₁+c₂⇒ 2(-k * -k + 0) = 6 + k⇒ 2(k² + 0) = 6 + k⇒ 2k² = 6 + k⇒ -2k² + k + 6 = 0

2g₁g₂ + 2f₁f₂ = c₁ + c₂2(g₁g₂ + f₁f₂) = c₁+c₂⇒ 2(-k * -k + 0) = 6 + k⇒ 2(k² + 0) = 6 + k⇒ 2k² = 6 + k⇒ -2k² + k + 6 = 0⇒ 2k² - k - 6 = 0

2g₁g₂ + 2f₁f₂ = c₁ + c₂2(g₁g₂ + f₁f₂) = c₁+c₂⇒ 2(-k * -k + 0) = 6 + k⇒ 2(k² + 0) = 6 + k⇒ 2k² = 6 + k⇒ -2k² + k + 6 = 0⇒ 2k² - k - 6 = 0⇒ 2k² + 3k - 4k - 6 = 0

2g₁g₂ + 2f₁f₂ = c₁ + c₂2(g₁g₂ + f₁f₂) = c₁+c₂⇒ 2(-k * -k + 0) = 6 + k⇒ 2(k² + 0) = 6 + k⇒ 2k² = 6 + k⇒ -2k² + k + 6 = 0⇒ 2k² - k - 6 = 0⇒ 2k² + 3k - 4k - 6 = 0⇒ k(2k + 3) - 2(2k + 3) = 0

2g₁g₂ + 2f₁f₂ = c₁ + c₂2(g₁g₂ + f₁f₂) = c₁+c₂⇒ 2(-k * -k + 0) = 6 + k⇒ 2(k² + 0) = 6 + k⇒ 2k² = 6 + k⇒ -2k² + k + 6 = 0⇒ 2k² - k - 6 = 0⇒ 2k² + 3k - 4k - 6 = 0⇒ k(2k + 3) - 2(2k + 3) = 0⇒ (k - 2)(2k + 3) = 0

2g₁g₂ + 2f₁f₂ = c₁ + c₂2(g₁g₂ + f₁f₂) = c₁+c₂⇒ 2(-k * -k + 0) = 6 + k⇒ 2(k² + 0) = 6 + k⇒ 2k² = 6 + k⇒ -2k² + k + 6 = 0⇒ 2k² - k - 6 = 0⇒ 2k² + 3k - 4k - 6 = 0⇒ k(2k + 3) - 2(2k + 3) = 0⇒ (k - 2)(2k + 3) = 0⇒ k = 2 , -3/2 (ans.)