find the center and radius of the circle with equation: x^2+y^2+6x-4y-15=0
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x^2 + y^2 + 6x - 4y - 15 = 0
First we need to rewrite the equation using the standard form for the circle:
( x-a)^2 + ( y-b)^2 = r^2 such that:
(a,b) is the center and r is the radius:
To rewrite the equation we need to complete the square:
==> (x^2 + 6x) + (y^2 - 4y) = 15
==> (x^2 + 6x + 9 -9 ) + ( y^2 - 4y -4 + 4) = 15
==> (x + 3)^2 - 9 + ( y-2)^2 + 4 = 15
==> (x+3)^2 + (y-2)^2 = 15 - 4 + 9
==> (x+3)^2 + (y-2)^2 = 20
Then the center is ( -3, 2) and r= sqrt20= 2sqrt5
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If a circle has the centre at (h,k) and a radius r, then the equation of the circle is given by:
(x-h)^2+(y-k)^2 = r^2.....(1).
Therefore we try bringing the given circle to the above form:
x^2+y^2+6x-4y -15 = 0
We rearrange add 15:
x^2+6x +y^2-4y = 15.
We make x^2+6x a perfect square by adding 3^2 and y^2-4y a perfect square by adding 2^2. Similarly, we add the same 3^2 and 2^2 to the other side of the equation also to balance.
x^2+6x+3^2+y^2-4y+2^2 = 15+3^2+2^2.
(x+3)^2 +(y-2)^2 = 15+9+4 = 28.
Therefore (x+3)^2+(y-2)^2 = (sqrt28)....(2)
Now the equations at (1) and (2) identically equal if and only if h= -3, k= 2 and r = sqrt28
Therefore the centre of the circle is (-3,2) and radus = sqrt28 = 2sqrt7.
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