The angle between the circles x2 + y2 + 4x + 2y + 1 = 0 and x2 + y2 - 2x + 6y - 6 = 0 is

Question

The angle between the circles x2 + y2 + 4x + 2y + 1 = 0 and x2 + y2 - 2x + 6y - 6 = 0 is (A)  (π/6) (B)  (π/3) (C)  (π/2) (D)  cos-1 (7/16)

Tardigrade Admin · Accepted Answer

Let θ be the angle of intersection of two given circles having centres (-g, -f ) = (-2, -1) and (-g1, -f1) = (1, -3) respectively, Then cosθ =|(-2 g g1 - 2 f f1 +c +c1/2√g2 +f2 -c √g12 + f12 -c1) | = |(-2(-2) -2(3)+1-6/2×√4 × √16)| =|(4-6-5/16)| = (7/16) ⇒ θ = cos-1 ((7/16))

Q. The angle between the circles $x^{2} + y^{2} + 4 x + 2 y + 1 = 0$ and $x^{2} + y^{2} - 2 x + 6 y - 6 = 0$ is

$\frac{π}{6}$

$\frac{π}{3}$

$\frac{π}{2}$

$cos^{- 1} \frac{7}{16}$

Q. The angle between the circles x2+y2+4x+2y+1=0 and x2+y2−2x+6y−6=0 is

6π​

3π​

2π​

cos−1167​

Q. The angle between the circles $x^{2} + y^{2} + 4 x + 2 y + 1 = 0$ and $x^{2} + y^{2} - 2 x + 6 y - 6 = 0$ is

$\frac{π}{6}$

$\frac{π}{3}$

$\frac{π}{2}$

$cos^{- 1} \frac{7}{16}$