Core 1 Mathematics circles

I'm really stuck on this question as I don't know where to start. If anyone could give me a hint on the first bit of working, I'm sure I'd be able to figure it out from there.
Thanks in advance!

The points P(a,b) and Q(c,d) are at the ends of a diameter of a circle. Show that the equation of the circle is (x-a)(x-c) + (y-b)(y-d) = 0

Reply 1

Megan_90

Original post by Ferdiee

Hi,

Have a look at this attachment! :smile:

12102756773_503962d26e_o.gif8.6KB

Reply 2

Ferdiee

Original post by Megan_90

Hi,

Have a look at this attachment! :smile:

Hey Megan, really helpful for that thanks a lot! I'm still confused where the x and y terms have gone after you've completed the square :confused:

, can you explain that?

Reply 3

Megan_90

Original post by Ferdiee

Hey Megan, really helpful for that thanks a lot! I'm still confused where the x and y terms have gone after you've completed the square :confused:

, can you explain that?

diameter = √[(a-c)² + (b-d)²]
therefore, the radius r = (½)√[(a-c)² + (b-d)²]
r² = [(a-c)² + (b-d)²]/4

Center of circle is the midpoint of PQ, or ((a+c)/2, (b+d)/2)
The equation of the circle is
(x - (a+c)/2)² + (y - (b+d)/2)² = [(a-c)² + (b-d)²]/4, x and y doesn't disappear it's just the way we are suppose to it work out, and what we are doing is using the coordinates given to work out that equation so we need to use the given coordinates only, it's the equation that contains the x and y. and we are simplifying it when u complete the square btw. Good question though!