(2,-1) [center of given circle]
d2 = 16
r12 = 12
=> r22 = 4
=> (x-2)2+(y-3)2=4
forgiv calc. errors
Find the eq. of the circle which cuts orthogonally the circle x2+y2-4x+2y-7=0 having the centre at (2,3)
I tried to use the relation d2=r12+r22 but then i got a quadratic eq. which was really wrong.
Help me.
(2,-1) [center of given circle]
d2 = 16
r12 = 12
=> r22 = 4
=> (x-2)2+(y-3)2=4
forgiv calc. errors
This is quite simple.
the equation of the circle is
(x-2)2 + (y-2)2 = r2
now expand it and then use the condition for orthogonal circles. You get r=2