Question

A circle has its center at (1, 4) and a radius of 2 units. What is the equation of the circle?

(x + 2)2 + (y + 4)2 = 2

(x - 1)2 + (y - 4)2 = 4

(x + 1)2 + (y - 4)2 = 4

(x - 1)2 + (y - 4)2 = 2

Answer 1

@ganeshie8

Answer 2

one sec

Answer 3

ok :D

Answer 4

[x-1]^2+[y-4]^2=4

Answer 5

@Dbzfan836 how ?

Answer 6

I used geogbra software, @Gabylovesyou

Answer 7

*Geogebra

Answer 8

ohhhhhhhhhhh ok ....

Answer 9

So basically all we're looking at here is a fancy form of the pythagorean theorem.

x^2+y^2=r^2



So that's the regular, general circle centered at the origin. Now,...

Answer 10

All we're saying is, let's just leave the radius constant and any x and y can vary as we move around the circle. For example, if we want the point right here at the top of the circle, we know that's when x=0
then we know that by plugging into our equation that: \[\Large 0^2+y^2=r^2\] Well all that says is that y=r. Which makes sense I hope?

Answer 11

yes :D

Answer 12

Ok awesome, so now let's try to move the circle somewhere and see what happens!

That looks about like a circle with a radius of 3. So let's just pretend that's what it is and that we've shifted it to the right by 1. Fair enough.

So now if we imagine doing the same thing we did last time, we want the x component to be zero when y is the length of the radius: That is to say: \[\Large y=3\]when\[\Large x=1\]So if we try to plug these suckers into this equation, things get horribly broken. POR QUE!? \[\Large 1^2+3^2 \ne 3^2\] We have to make that x term equal zero for the circle to make sense. So we conveniently subtract it off: \[\Large (1-1)^2+3^2=3^2\] That looks much better. So now let's ask ourselves, what's negative and what's not?

Well we know that top point is at the location (1,3) so it looks like the 1 that is being subtracted is what we're shifting by.\[\Large (x-1)^2+y^2=3^2\] Play around with it a little. In fact you will see that this is a fairly good general rule for shifting all other things around like if you have a parabola or something and want to shift it to the right then you subtract the amount you want to move.

Answer 13

See, now we can try to use this equation to see what happens when y=0. It would seem to me that this point we get is So that means if we plug in y=0 we should have x=4. But does the math work out? If we guessed right in the last thing, it should.

\[\Large (x-1)^2+0^2=3^2\]

Test it out!

Now you should feel fairly comfortable with the idea that: \[\Large (x-h)^2+(y-k)^2=r^2\] where (h,k) is just the center of the circle. Since things are round and symmetrical, I just used the same reasoning for moving things on the x-axis to moving things on the y-axis.

I hope that clears up a little bit how to reason out why this is how this should look?

Answer 14

yes thanks so much !