find the equation of the circle concentric with the circles x^2 + y^2 -4x +6y-3=0 and of double its (1) circumference (2) area

^{2}+ y

^{2}- 4x + 6y -3 = 0

So changing it as standard equation, as (x - h)

^{2}+ (y -k)

^{2}= a

^{2}, here ( h , k) are centre of the circle and a is the radius.

x

^{2}+ y

^{2}- 4x + 6y -3 = 0

x

^{2}+ y

^{2}- 4x + 6y = 3

Adding 4 and 9 on both sides we get,

x

^{2}+ y

^{2}- 4x + 4 + 6y + 9 = 3 + 4 + 9

(x - 2)

^{2}+ (y + 3 )

^{2}= 4

^{2}(i)

So the centre is ( 2 ,-3) and Radius is 4.

1) Circumference is double, hence radius is double, but centre remains same as the circle are concentric

So the Radius will be 8

So changing the radius value of equation (i) to 8, we get

(x - 2)

^{2}+ (y + 3 )

^{2}= 8

^{2}

Hence on expanding we get x

^{2}+ y

^{2}- 4x + 6y + 4 + 9 = 64

x

^{2}+ y

^{2}- 4x + 6y - 51 = 0 , is the equation of the circle.

2) Area is double

Initially the area was πr

^{2}= π(4)

^{2 }= 16π

As the area is double , so the area is 32π

For the area of 32π, we have radius of πr

^{2}= 32π or r = $\sqrt{32}$

So the equation is (x - 2)

^{2}+ (y + 3 )

^{2}= $(\sqrt{32}{)}^{2}$

Hence x

^{2}+ y

^{2}- 4x + 6y + 4 + 9 -32 = 0

x

^{2}+ y

^{2}- 4x + 6y - 19 = 0

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