The Student Room Group

Reply 1

(x+3)2 = x2+6x+9
(y-4)2 = y2-8x+16

(x+3)2+(y-4)2 = x2+6x+y2-8x+25

Therefore (x+3)2+(y-4)2-4 = x2+6x+y2-8x+21

So (x+3)2+(y-4)2 -4 = 0 means (x+3)2+(y-4)2 = 22

Therefore centre at (-3,4) with radius 2.

For part b, try drawing a line through the origin and the centre of the circle and seeing where it intersects the circle.

Reply 2

Ty - heres the next lol

Circle C1 has the equation x2 + y2 − 2ay = 0, where a is a positive constant.

Find the coordinates of the centre and the radius of C1.
Circle C2 has the equation x2 + y2 − 2bx = 0, where b is a constant and b > a.

Reply 3

When the equation of a circle is in the form:

x2 + y2 + 2gx + 2fy + c

Centre (-g, -f)

Radius = Sqrt(g2 + f2 -c)

Reply 4

For the first question part b:
If the point P lies on the circle its distance must be equal to the radius of the circle. Therefore maximum distance is root4=2.