# 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

Equation of the circle is given as x2 + y2 - 4x + 6y -3 = 0
So changing it as standard equation, as (x - h)2 + (y -k)2 = a2 , here ( h , k) are centre of the circle and a is the radius.
x2 + y2 - 4x + 6y -3 = 0
x2 + y2 - 4x + 6y = 3
Adding 4 and 9 on both sides we get,
x2 + y2 - 4x  + 4 + 6y + 9 = 3 + 4 + 9
(x - 2) 2+ (y + 3 )2 = 42  (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 = 82
Hence on expanding we get x2 + y2 - 4x + 6y + 4 + 9 = 64
x2 + y2 - 4x + 6y -  51 = 0 , is the equation of the circle.

2) Area is double
Initially the area was πr2 = π(4)2 = 16π
As the area is double , so the area is 32π
For the area of 32π, we have radius of πr2 = 32π or r = $\sqrt{32}$
So the equation is (x - 2) 2+ (y + 3 )2$\left(\sqrt{32}{\right)}^{2}$
Hence x2 + y2 - 4x + 6y + 4 + 9 -32 = 0
x2 + y2 - 4x + 6y - 19 = 0

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