Circle geometry

hey there, i have posted this question before, but I still don't full understand.

A detailed explanation would be so so helpful! thanks

a circle passes through the points (0,2) and (0,8) and it's centre lies on the line x=4.

find equation of the circle.

Consider: (4,y) as the centre of the circle - you also know that two points lie on the circle.
What can you tell me about the lengths of the line from the centre to the points on the circle?

they are equal.

but what do I do with this?

If the equation of the circle takes form (x-a)^2+(y-b)^2=r^2 where r is the radius, and (a,b) the centre, we know that the x-coordinate for the centre is 4 so we can write centre=(4,b) and (x-4)^2+(y-b)^2=r^2.

We have two variables we need to find (b and r) and two sets of coordinates: (0,2) and (0,8). Plug these in to get two equations:

(0-4)^2+(2-b)^2=r^2
(0-4)^2+(8-b)^2=r^2

Inspection makes it immediately obvious that (2-b)^2=(8-b)^2. Solve this out and the b^2 terms will cancel and you'll be left with 4-4b=64-16b; add 16b to both sides and take away 4 and you get 12b=60, so b=5. Now back-substitute into either of those two equations we came up with originally and solve and you'll find r^2=25 (r=5) so our equation is (x-4)^2+(y-5)^2=25.

Hope this helps.

Full solutions are against forum guidelines . . .

If you read this thread in some detail: http://www.thestudentroom.co.uk/showthread.php?t=415707&page=2&page=2 (start from page 1) you will see that full solutions if reasoning is given behind the steps are beneficial to the OP.

If the equation of the circle takes form (x-a)^2+(y-b)^2=r^2 where r is the radius, and (a,b) the centre, we know that the x-coordinate for the centre is 4 so we can write centre=(4,b) and (x-4)^2+(y-b)^2=r^2.

We have two variables we need to find (b and r) and two sets of coordinates: (0,2) and (0,8). Plug these in to get two equations:

(0-4)^2+(2-b)^2=r^2
(0-4)^2+(8-b)^2=r^2

Inspection makes it immediately obvious that (2-b)^2=(8-b)^2. Solve this out and the b^2 terms will cancel and you'll be left with 4-4b=64-16b; add 16b to both sides and take away 4 and you get 12b=60, so b=5. Now back-substitute into either of those two equations we came up with originally and solve and you'll find r^2=25 (r=5) so our equation is (x-4)^2+(y-5)^2=25.

Hope this helps.