Circle Equation Question (1 Viewer)

[frenzal_dude]

A circle touches the +ve x and y axes and also passes through the point (1,2).
Determine the centre and radius of all such circles.

The question says 'of all such circles' but shouldn't there just be only 1 circle with this description? I tried to work it out but I didn't know to get exact answers with no variables.

 

[MetroMattums]

Does touching mean the x and y axes are tangential to the circle?

 

[fullonoob]

there are two such circles because if the circle equation passes through (1,2) it could be facing upwards or downwards (one is concaving down and one is concaving up). I haven't worked it out yet (in school ) but i did x^2 +y^2 = r^2, sub in points; r = root 5. But since centre is not at origin, draw triangle to find centres or distance formula etc. using r = root 5.

 

[nikkifc]

A circle touches the +ve x and y axes and also passes through the point (1,2).
Determine the centre and radius of all such circles.

The question says 'of all such circles' but shouldn't there just be only 1 circle with this description? I tried to work it out but I didn't know to get exact answers with no variables.

I'm really starting to get concerned, since you're a maths tutor...

and yes, there is only one such described circle...

 

[frenzal_dude]

I'm really starting to get concerned, since you're a maths tutor...

and yes, there is only one such described circle...

You just want to express your concern about some random person you don't know not being good enough to be a tutor? Or do you actually have an answer to the question?

BTW you're wrong, there are 2 circles, one little one, and one larger one which is flipped.

 

[Drongoski]

Since circle touches the x & y axes and passes thru the point (1,2) it lies in the 1st quadrant
and has centre (h.h) and radius h.








If you live in my area and need a good(????) personal maths tutor, pm me. Not cheap though.

 

[frenzal_dude]

This is actually an easy question, can't believe noone's given the answer yet:

If the circle touches the x and y axes, then the centre is going to be (h,k) where h=k.

The point where it touches the y axis is (0,k), the point where it touches the axis is (h,0), subbing these along with (1,2) into (x-h)^2 + (y-k)^2 = r^2 gives you h=r=k.

Then just sub (1,2) into (x-h)^2 + (y-h)^2 = h^2 (since h=r=k) will give you h=5 or 1.

So if h = 5, you have a circle centre at (5,5) and r=5,

If h=1 you have a circle centre at (1,1) and r = 1 !

 

[frenzal_dude]

thanks Drongoski, haha u beat me by 2 mins!

 

[Drongoski]

thanks Drongoski, haha u beat me by 2 mins!

you are welcome.

 

[fullonoob]

oh jeesus thought it said pass through x y axes

 

[nikkifc]

You just want to express your concern about some random person you don't know not being good enough to be a tutor? Or do you actually have an answer to the question?

Wow, look at him fire up. I certainly wouldn't want to be tutored by some person who is NOT ARROGANT AT ALL, like yourself. A very touche person indeed.

BTW you're wrong, there are 2 circles, one little one, and one larger one which is flipped.

Well obviously I didn't try the question. But you're supposed to be a tutor that has Extension 1 secured under their belt and not posting on BOS every time your student has a question. Also note that you're about 5 years older than myself and thus you would be comparing apples with oranges if you were to compare yourself with me.

And yes I had an idea how to do it, as Drongoski noted, the radius was equal to each of the coordinates of the centre.