Check Eqn of Circle with diameter end points(1,2) and (3,-6)

[aisha]

Check please Eqn of Circle with diameter end points(1,2) and (3,-6)

Find the equation of a circle with a diameter that has end points P(1,2) and Q(3,-6)

I know the equation for a circle in general form is $$x^2+y^2=r^2$$

First I found the distance between the two points given because this also equals the diameter.

I used point P as x1,y1 and Q as x2,y2 and subbed these points into the distance equation.

I got Distance=Diameter=sqrt(68) I took half of this to get the radius. 5.83 and then I put this into the equation for a circle and squared it to get 33.99

My final equation of a circle for this question is $$x^2+y^2=34$$

Can someone please check my answer and tell me if I went wrong anywhere?

 

[F|234K]

if the equation is $$x^2+y^2=34$$, then the centre is (0,0). but according to your diameter, the centre should not be (0,0). btw, the radius is 4.1231...

i think the equation is (x-2)^2+(y+2)^2=17

 

[aisha]

um using the points I found the midpoint to be (1,-4) does that mean this equation should be written as

$$(x-1)^2 + (y+4)^2 = 34$$

 

[F|234K]

no...$$(x-1)^2+ (y+4) ^2= 34$$

 

[F|234K]

lol...i graphed both equations (mine and yours) in my graphic calculator...both are wrong...

 

[aisha]

I was just going to write the answer that you wrote I remember I standard form, are you sure it is wrong? Maybe only the 34 part is wrong I had to do some rounding.

(x-2)^2+(y+2)^2=17 is this one right?

 

[F|234K]

acutally, i really think your radius is wrong, sqrt 68 then divide by 2 is 4.1231. i also think that the transformation part is right , but it turns out to be wrong in my calculator

 

[F|234K]

"(x-2)^2+(y+2)^2=17 is this one right?"

i think so. but my graphic calc says i am wrong

 

[aisha]

yes my radius was wrong now I think the equation is

$$(x-1)^2 + (y+4)^2 = 17$$ what do you think? this seems right

the coordinates of the center will be the midpoint of the two points which is (1,-4)

 

[F|234K]

nope, it's even more wrong then mine

 

[F|234K]

why you think the radius is (1,-4)

i think its (2,-2)

 

[aisha]

I don't think that is the radius I think that is the center point (h,k)

The question says find the equation of a circle the diamter has end points P(1,2) and Q(3,-6) This is the diameter the middle of the diamter will be the center that's why I am taking the midpoint as (1,-4)

[(x2-x1)/2, (y1-y2)/2]

 

[F|234K]

think about this, the diameter given is sloped, thus the centre of the diameter wouldnot have the same x value as one of its end points...

 

[F|234K]

i think my graphic calc is wrong (since i rememebr my teacher told me that he might accidently changed my program). just copy my equation.

 

[aisha]

ok tell me where you got ur point (2,-2)?

 

[Data]

The midpoint formula is $$\left( \frac{x_1+x_2}{2}, \ \frac{y_1+y_2}{2}\right)$$, not $$\left( \frac{x_2-x_1}{2}, \ \frac{y_2-y_1}{2} \right)$$.

 

[F|234K]

so is my equation wrong data? because i can't find out where i got wrong.

 

[Data]

I believe $(x-2)^2 + (y+2)^2 = 17$ is correct, yes~

By the way, there's no need to convert anything to decimal form for this question, or for most others. Your radius is $r$ in

$$r^2 = \left(\frac{\sqrt{68}}{2}\right)^2 = \frac{68}{4} = 17.$$

 

[dextercioby]

Because the coordinates of the center of the circle must be in certain relation with the ones of the ends of the diameter.Use the symmetry of the circle to get the correct equation as

$$(x-2)^{2}+(y+2)^{2}=17$$

U can check that 2 points are on that circle &,moreover,the radius is $\sqrt{17}$...

Daniel.

 

[aisha]

hey thanks so much, my midpoint formula was wrong in my online text no wonder I wasnt getting the point (2,-2) but now everything is fine.

 

[HallsofIvy]

Arithmetic: halfway between 1 and 3 is (1+3)/2= 2, halfway between 2 and -6 is (2-6)/2= -2.

By the way, the difference between r²= 34 and r²= 17 is surely not "round off error"! If d²= 68 and r= d/2 then r²= d²/4= 68/4= 17.

 

[maverick280857]

Do you know that the angles in the same segments are equal? (This is geometry). A useful corollary of this is that the angle between the lines joining the ends of a diameter is always 90 degrees for the circle. And the point of intersection is any point on the circle. So if the ends of a diameter are $(x_1,y_1)$ and $(x_2,y_2)$ then the equation of the circle is $(x-x_1)(x-x_2)+(y-y_1)(y-y_2) = 0$ and this is fairly straightforward coordinate geometry (I'll let you prove it).

EDIT: To be fair, this is . close to what Hurkyl said, in general terms.