# Find the centre of the circle that passes through points A(-7,4) B(-4,5) and C (O,3)

embizze | High School Teacher | (Level 1) Educator Emeritus

Posted on

The circle should pass through the three points (-7,4),(-4,5) and (0,3)

Call the center of the circle P. Then PA=PB=PC since all radii of a circle are congruent.

PA=PB implies that P lies on the perpendicular bisector of AB; similarly P lies on the perpendicular bisector of BC and AC.

The perpendicular bisector of a segment goes through the midpoint of the segment, and the slope is the opposite reciprocal of the slope of the segment.

(i) The `_|_` bisector of AB: The midpoint of AB is `((-7+-4)/2,(4+5)/2)=(-11/2,9/2)` . The slope of AB is `1/3` , so the slope of the `_|_` bisector is -3. Then the equation for the `_|_` bisector of AB is `y-9/2=-3(x+11/2)==> y=-3x-12`

(ii) The `_|_` bisector of BC: The midpoint of BC is `((-4+0)/2,(5+3)/2)=(-2,4)` The slope of BC is `-1/2` , so the slope of the `_|_` bisector is 2.

Then the equation of the `_|_` bisector is `y-4=2(x+2)==>y=2x+8`

(iii) The perpendicular bisectors meet when `-3x-12=2x+8==>x=-4` and `y=0`

---------------------------------------------------------------

The center of the circle going through the given points is (-4,0)

---------------------------------------------------------------

** You could check by finding the `_|_` bisector of AC: the midpoint is (-3.5,3.5) and the slope of the `_|_` bisector will be 7 giving you `y-3.5=7(x+3.5)==> y=7x+28`

Let `-3x-12=7x+28==>x=-4,y=0` as before.

This is the circumcircle of triangle ABC, and P is the cicumcenter.

Sources:

vaaruni | High School Teacher | (Level 1) Salutatorian

Posted on

Let  O(x,y)  be the center of the circle. Since the circle passes through the poins A(-7,4) , B(-4,5) and C(0,3) therefore OA, OB and OC will be the radius of the circle and hence OA= OB= OC. Now we wind the distance(length) OA , OB and OC  using distance formula for distance between two points A(x1,y1) and B(x2,y2)                                         AB ^2 = (x1 - x2)^2 + ( (y1 - y2) ^2

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

OB^2 = (x - (-4))^2 + (y - 5)^2 = (x + 4)^2 +(y - 5)^2  ------(2)

OC^2 =  (x - 0 )^2 + (y - 3)^2  = x ^2 + (y - 3)^2    ----------(3)

Since OA^2 = OB^2 = OC^2 , there fore equating equations(1) & (3):

(x + 7)^2 + (y - 4)^2 = x^2 + (y - 3)^2

Or,  x^2 +14x+ 49 + y^2- 8y+ 16 = x^2+ y^2- 6y+ 9

Or,  14x- 8y 6y = -49 - 16 +9

Or,  14x 2y = 56

Or,   7x - y = -28  --------(4)

Now equating eqn. (2) and (3) :

(x + 4)^2 + (y - 5)^2 = x^2 + (y - 3)^2

Or, x^2+ 8x+ 16+ y^2- 10y+ 25 = x^2 + y^2 - 6y + 9

Or, 8x- 10y +6y = -16- 25+ 9

Or, 8x- 45 = -32

Or,  2x - 7y = -8  --------(5)

Now solving eqn. (4) and (5) simultaneously :

7x - y = -28

2x - y = -8                                                                                   -     +      +

----------------

5x   =    -20

therefore   x = -4 ,  Now substituting value of x in equation (5)

We get  2x - y = -8  , i.e.    2.(-4) - y = -8

Or,   -8 - y = -8

Or,   -y = -8 + 8 = 0

Thus we got x = -4 and  y = 0

Hence, The radius of the circle = ( -4 , 0 )    Answer