# The perpendicular bisector of a chord of a circle passes through its centre.Find the centre passing through the points P(5,7), Q(7,1) and R(-1,5) by finding the perpendicular bisectors of PQ and QR...

The perpendicular bisector of a chord of a circle passes through its centre.

Find the centre passing through the points P(5,7), Q(7,1) and R(-1,5) by finding the perpendicular bisectors of PQ and QR and solving them simultaneously

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We are asked to find the center of a circle that passes through the points P(5,7),Q(7,1) and R(-1,5):

(1) First we find the equation of the perpendicular bisector of PQ:

To find the equation of a line we need a point and the slope.

(a) The point we require is the midpoint of `bar(PQ)` . The coordinates of the midpoint of a segment are the means of the coordinates of the endpoints; if the endpoints are (a,b) and (c,d) then the midpoint is `((a+c)/2,(b+d)/2)` . Here the midpoint is (6,4).

(b) The slope of the line through P and Q is `m_(bar(PQ))=(7-1)/(5-7)=-3` . The perpendicular bisector is perpendicular to this line so its slope is the opposite reciprocal. The slope of the perp. bisector is `m=1/3`

(c) We have a point (6,4) and a slope `m=1/3` . The equation of the line is `y-4=1/3(x-6)==>y=1/3x+2`

(2) We also need the equation of the perpendicular bisector through `bar(QR)` . We proceed as in (1):

(a) The midpoint is (3,3)

(b) The slope through Q and R is `m_(bar(QR))=(5-1)/(-1-7)=-1/2`

The slope of the perpendicular bisector is 2.

(c) We have the point (3,3) and the slope m=2. The equation of the perpendicular bisector is y-3=2(x-3) ==> y=2x-3

(3) We now solve the system `y=1/3x+2,y=2x-3` to find the center. We can use substitution:

`1/3x+2=2x-3`

`x+6=6x-9`

`5x=15`

x=3 ==> y=3

**The center of the circle is (3,3).**

The graph:

Given point P(x1,y1) = (5,7), Q(x2,y2) = (7,1), R(x3,y3) = (-1,5)

Let the co-ordinate of the center O be (x,y) and mid point of line

PQ be S and mid point of QR be T. We find the co-ordinates of the mid point (S) of the line PQ and mid point (T) of the line QR using section formula for mid point

X - coordinate of 'S' -> x = (x1+x2)/2

x = (5+7)/2 = 12/2 = 6

Y- coordinate of 'S' -> y = (y1+y2)/2

y = (7+1)/2 = 8/2 = 4

We found S(x,y) = (6,4)

Similarly we will find the co-ordinates of the mid point (T) of the line QR

X - coordinate of 'S' -> x = (x1+x2)/2

x = (7-1)/2 = 6/2 = 3

Y- coordinate of 'S' -> y = (y1+y2)/2

y = (1+5)/2 = 6/2 = 3

We found T(x,y) = (3,3)

Line OS and OT are the perpendicular bisector of the lines PQ and QR Joining OQ we get two right angled triangles OST and OTQ respectively

Next we will find the length of the lines OS, OT and OQ using

Distances formula [ sqrt((x1-x2)^2+ (y1-y2)^2) ]

OS^2 = (x-6)^2+(y-4)^2 , ST^2= (6-7)^2+(4-1)^2= 1+9 = 10

OT^2= (x-3)^2+(y-3)^2 , QT^2=(7-3)^2+(1-3)^2= 16+4 =20

OQ^2= (x-7)^2+(y-1)^2

We know that OS^2+ST^2 = OQ^2 [ Pythagoros theorem ]

(x-6)^2+(y-4)^2 + 10 = (x-7)^2+(y-1)^2

=> x^2-12x+36+y^2-8y+16+10 = x^2-14x+49+y^2-2y+1

=> x^2 +y^2-12x-8y+62= x^2+y^2-14x-2y+50

=> -12x+14x-8y+2y=50-62

=> 2x – 6y = -12

=> x – 3y = -6 ------(1)

Similarly using OT^2 +QT^2 = OQ^2

(x-3)^2+(y-3)^2+ 20 = (x-7)^2+(y-1)^2

=> x^2-6x+9+y^2-6y+9 +20 = x^2-14x+49+y^2-2y+1

=> x^2+y^2-6x-6y+38 = x^2+y^2-14x-2y+50

=> -6x+14x-6y+2y = 50-38

=> 8x-4y = 12

=> 2x – y = 3 ------(2)

Solving equations (1) and (2) simultaneously :-

2x – y = 3

2x – 6y = -12 [ multiplying Eq. (1) by 2 ]

=> (2x-y)-(2x-6y) = 3-(-12) [ eq.(1) – eq.(2) ]

=> 2x-2x -y+6y = 3+12

=> 5y = 15

=> y =3

Substituting value of y=3 in equation (1)

x – 3y = -6

=> x – 3*3 = -6

=> x = -6 + 9 = 3

x = 3, y = 3

Hence **x = 3, y = 3**