# find the center and radius of the circle with equation: x^2+y^2+6x-4y-15=0

hala718 | High School Teacher | (Level 1) Educator Emeritus

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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

neela | High School Teacher | (Level 3) Valedictorian

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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

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.