The equation of a circle is x^2-6x+y^2+4y=12. Complete the square to find the centre and radius of the circle.
When we complete the square, we're looking for an equation that looks like (x-a)2 + (y-b)2 = r2, where (a,b) is the centre and r is the radius. When we expand brackets using FOIL, the x part is (x2-2ax+a2). This means the term with x in it has a coefficient of 2a. Therefore halving it will give us a! So let's apply this to our equation. The x term has a coefficient of -6 and the y term has a coefficient of +4. Dividing these by 2 we get -3 and 2. So the centre is (-3,2)! Now since (-3)2=9 and (2)2=4 we can add these on two complete the squares: x2-6x+9-9 + y2+4x+4-4=12 (x-3)2-9 + (y+2)2 -4 = 12 (x-3)2 + (y+2)2 = 25 (x-3)2 + (y+2)2 = 52 So the radius is 5!
HA
Related Maths A Level answers
All answers ▸Given that y = 16x + x^-1, find the two values of x for which dy/dx = 0
Starting from the fact that acceleration is the differential of velocity (dv/dt = a) derive the SUVAT equations.
Differentiate y=x^3+ 7x-ln(2)
g(x) = ( x / (x+3) ) + ( 3(2x+1) / (x^2 + x - 6) ). Show that this can be simplified to: g(x) = (x+1) / (x-2).