A) \( \Large x^{2}+y^{2}+2x-2y-62=0 \) |
B) \( \Large x^{2}+y^{2}-2x+2y-62=0 \) |
C) \( \Large x^{2}+y^{2}-2x+2y-47=0 \) |
D) \( \Large x^{2}+y^{2}+2x-2y-47=0 \) |
C) \( \Large x^{2}+y^{2}-2x+2y-47=0 \) |
The given equation of diameters are
\( \Large 3x-4y-7 = 0 \) ...(i)
and \( \Large 2x-3y-5 = 0 \) ...(ii)
On solving equations (i) and (ii) we get
x = 1 and y = -1
Centre of circle is \( \Large \left(1,\ -1\right) \)
Let r be the radius of circle then
\( \Large \pi r^{2} = 49 \pi => r = 7\ unit \)
Equation of required circle is
\( \Large \left(x-1\right)^{2}+ \left(y+1\right)^{2} = 49 \)
=> \( \Large x^{2}+y^{2}-2x+2y+1+1 = 49 \)
=> \( \Large x^{2}+y^{2}-2x+2y-47 = 0 \)