Question

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

Accepted Answer

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.

Answer

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.

Answer

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

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

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