The centre of the circle passing through the points (8,12),(11,3) and (0,14) is​

Let the equation of the circle be x2+y2+2gx+2fy+c=0 (i)

Then its centre will be O=(−g,−f). 

It passes through A=(x1,y1)=(8,12),B=(x2,y2)=(11,3) and C=(x3,y3)=(0,14). 

Substituting (x,y)=(x1,y1)=(8,12),(x2,y2)=(11,3) and (x3,y3)=(0,14) in equation (i) and simplifying, we get 

16g+24y+c=−208 (ii),

22g+6f+c=−130 (iii)

and 28f+c=−196 (iv) 

Eliminating c from (ii) and (iii),

g=3f+13 (v)

Again eliminating c from (iii) and (iv), we have

g=f+3 (vi)

∴ From (v) and (vi), we get

3f+13=f+3

Putting f=−5 in (vi),  have

g=−2

∴ Centre O=(−g,−f)=(2,5)

(2,5)

#CarryOnLearning