Question

find the paramatric equation of the circle x2+y2-4x+6y-3=0​

Answer 1

Let us assume a equation of circle
(x-a)^2+(y-b)^2=r^2
Since the above circle is concentric with given circle i.e.,x^2+y^2-4x-6y-9=0
The center of given circle will be (d,e)
d=-(-4/2)
e=-(-6/2)
(d,e)=(2,3)
The centre of assumed circle will also be (2,3) 
Because both are concentric circles .
Therefore , we can say a=2,b=3
 And circle is also passing through (-4,-5) 
Therefore
(-4-2)^2+(-5-3)^2=r^2
(-6)^2+(-8)^2=r^2
36+64=r^2
r^2=100
r=√100
r=10
Equation of circle will be 
(x-2)^2+(y-3)^2=100



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