# Find the centre and radius of the circle:x^2 + y^2 + 8x + 6y= 0

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

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x^2 + y^2 + 8x + 6y= 0

We need to re-write the equation in the following format:

( x- a)^2 + ( y-b)^2 = r^2   such that:

(a,b) is the center of the circle and r is the radius:

Let us calculate:

We will re-arrange terms:

x^2 + 8x + y^2 + 6y = 0

We will complete the squares:

==> ( x+4)^2 - 16 + ( y+3)^2 - 9 = 0

==> ( x+4)^2 + ( y+3)^2 - 25 = 0

We will move 25 to the right side:

==> ( x+ 4)^2 + ( y+3)^2 = 25

==> ( x+ 4)^2 + ( y+3)^2 = 5^2

Now the equation is in the standard form:

Then the center of the circle is ( -4, -3) and the radius = 5

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have the equation of the circle given as x^2 + y^2 + 8x + 6y = 0.

Now, the equation of a circle with center (h, k) and radius r is (x − h)^2 + (y − k)^2 = r^2

So we convert x^2 + y^2 + 8x + 6y = 0 to the standard form.

x^2 + y^2 + 8x + 6y = 0

=> x^2 + 8x + y^2 + 6y = 0

=> x^2 + 8x + 16 + y^2 + 6y + 9 = 16 + 9

=> (x + 4)^2 + (y +3)^2 = 25 = 5^2

So the center is ( -4, -3) and the radius is 5.

The centre and radius of the circle x^2 + y^2 + 8x + 6y = 0 is ( -4, -3) and 5 respectively.

neela | High School Teacher | (Level 3) Valedictorian

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We know that the equationof a circle with centre (h,k) and radius ris given by:

(x-h)^2+(y-k)^2 = r^2......(1)

We now recast x^2+y^2+8x+6y = 0 in the form at (1) and equate the like terms:

X^2+8x = (x+4)^2 -4^2.

y^2+6y = (y+4)^2 -3^2

Therefore x^2+y^2+8x+6y = (x+4)^2+(y+3)^2 -4^2-3^2.

Therefore the equation x^2+y^2+8x+6y= 0 is the same as:

(x+4)^2+(y+3)^2 -25 = 0. Or

(x+4)^2+(y+3)^2 = 25 = 5^2.

{x-(-4)^2}^2 + {y- (-3)}^2 = 5^2.....(2).

Since eq (1) and eq(2) are same we can equate like terms:

{x-h)^2 = {x-(-4)}^2 gives h = -4.

Similarly, (y-k)^2 = {y-(-3)}^2  gives k= -3.

r^2= 5^2 gives  r= 5.

Therefore  the centre of the circle = (-4,-3) and the radius of the circle = 5.