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
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.