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# Circle Geometry watch

1. 1) Triangle PQR has vertices P(-2, -1) Q(-7, 4) R(-1, 1)

a) Show PR is perpendicular to QR.

Now for this I thought if i found the gradients of both lines using y2-y1/x2-x1 then by showing they both multipy to make -1.

But that gave me 8/5 x 3 which doesnt equal -1 so that doesnt show that they are perpendicular...so i think ive gone wrong here somewhere.

2) PQ is a diameter of a circle. Find an equation of the circle when:

a) P is (2, 0) and Q is (6,0)

With this i assumed that if i found the midpoint of the line then that would give me the centre of the circle which I could then place into the equation :
(x-a)^2 + (y-b)^2 = r^2

Then to find the radius i thought by finding the magnitude of the line then i could find the diameter and half it.

I got the answer to be (x-1)^2 + (y-3)^2 = 4 but the book tells me its
(x-4)^2 + y^2 = 4

Could someone attempt to explain where ive gone wrong on these two questions.

Thanks
2. 1a) mPR=((dy)/(dx))=((1--1)/(-1--2))=(2/1)=2
mQR=((dy)/(dx))=((1-4)/(-1--7))=(-3/6)=-(1/2)

If two lines are perpendicular ((m1)(m2))=-1

Let m1=mPR and m2=mQR. Therefore:

((m1)(m2))=((2)(-1/2))=-1 Hence PR is perpendicular to QR.

Q. E. D.

2a) P(2,0) Q(6,0) and PQ is the diameter of the circle. Find the mid-point to find the centre of the cirlce.

MP=(((x1+x2)/(2)),((y1+y2)/(2)))=(((2+6)/(2)),((0+0)/(2)))

=> = (((8)/(2)),((0)/(2)))=(4,0)

Cartesian equation of a circle:

(((x-a)^2)+((y-b)^2))=(r^2) where a and b are the x and y co-ordinates, respectively, of the centre of the circle and r is the radius of the circle.

a=4
b=0

r: The radius is the magnitude of the distance between one side of the diameter to the midpoint of the circle. Let us consider P(2,0). Since the co-ordinates of the centre are (4,0). We can tell that the radius is 4-2=2units.

Therefore the equation of the respective circle is:

(((x-4)^2)+((y-0)^2))=(2^2)

=>(((x-4)^2)+(y^2))=(4) Q. E. D.

Newton.
3. thanks alot.

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