Show that the equation
5y^2+3x^2+4 = 4x−2x^2−10y
represents a circle by rearranging it into the centre-radius form of the equation of a circle. State the coordinates of the centre and the radius of the circle.



... how do you do that?

Hey mate,

a circle with radius r, centered at h,k can be represented in cartesian coordinates by the equation,

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

obviously for your situation you have the 'expanded out' version of the above general equation and need to transform it into its classic form.

Note a task to dread, but can be painfully algebraically draining! it all centres around the concept you may have heard at school known as 'completing the square' which simply means that equations of the form
a^2 +/- 2ab + b^2 = (a +/- b)^2

i.e.

x^2 + 6x + 9 = (x + 6)^2

The above example is fairly straight forward to perform, however when the expression becomes more 'ugly' how does one complete the square and furthermore - can it be completed in full?

With all due respect, Im assuming that as a high school student you would not be interested into the governing mechanics of the following methodology so I will simply present the result,

x^2 + a*x + b = (x + (a/2))^2 + (b - (a/2)^2)

if we were to remove b from both sides we achieve

x^2 +a*x = (x + (a/2))^2 - (a/2)^2

we now have all the ammunition we require to complete the underlying question,

5y^2+3x^2+4 = 4x−2x^2−10y

First Step : Group all common variables together,
thus,
(5y^2 + 10y) + (5x^2 - 4x) + 4 = 0

here we have a problem, the coefficent of y^2, x^2 are not 1, as required, to overcome this divide by 5
thus,
(y^2 + 2y) + (x^2 - (4/5)x) + (4/5) = 0
Second Step : complete the squares for x, y separately using the formula from before
y^2 + 2y = (y + 1)^2 - 1
x^2 - (4/5)^2 = (x - (2/5))^2 - 4/25
Third Step, sub back into equation,
thus
( (y+1)^2 - 1 ) + ( (x - (2/5))^2 - 4/25 ) + 4/5 = 0
simplifying yields
(y + 1)^2 + (x - 2/5)^2 = 9/25
We now have the equation in a classical circle form
therefore
h = 2/5, k = -1 and r = sqrt(9/25) = 3/5

Hope this helps,

David