Find the angle of intersection of the circles `x^2+y^2-6x+4y+11=0andx^2+y^2-4x+6y+9=0`

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

B

Explanation

The equations of the circles are
`x^(2)+y^(2)-6x+4y+11=0` ...(i)
`x^(2)+y^(2)-4x+6y+9=0` ...(ii)
The centres of these circles are `C_(1) (3, -2) and C_(2)(2, -3)` respectively. Let `r_(1) and r_(2)` be their radii. Then,
`r_(1)=sqrt(9+4-11)=sqrt(2) and r_(2) = sqrt(4+9-9)=2`.

Also, `C_(1)C_(2)`
Suppose circles (i) and (ii) intersect at P. Then,
`cos angleC_(1)PC_(2)=(C_(1)P^(2)+C_(2)P^(2)-C_(1)C_(2)^(2))/(2C_(1)P.C_(2)P)=(2+4-2)/(2xxsqrt(2)xx2)=(1)/(sqrt(2))`
`rArr angle C_(1)PC_(2)=45^(@)`
Thus, the angle of intersection of the given circles is of `45^(@)`.