Question

If circles x² + y² + kx + 4y + 2= 0 and

2(x² + y²) – 4x – 3y + k = 0 intersect orthogonally, then k is equal to

• A. $\frac{10}{3}$
• B. $- \frac{10}{3}$
• C. $\frac{8}{3}$
• D. –$\frac{8}{3}$

Answer 1

Answer: (B)

$- \frac{10}{3}$

Answer 2

x² + y² + kx + 4y + 2 = 0 and

x² + y² – 2x – $\frac{3}{2}$y + $\frac{k}{2}$= 0

\ 2g₁ g₂ + 2f₁ f₂ = c₁ + c₂

2 $\left( \frac{k}{2} \right)( - 1) + 2(2)\left( - \frac{3}{4} \right) = 2 + \frac{k}{2}$

– k + (–3) = $\frac{4 + k}{2}$

– 2k – 6 = 4 + k

– 3k = 10 ̃ k = –$\frac{10}{3}$