Question

Find the centre of a circle passing through the points 6, d6 3, - 7 and 3, 3

Answer 1

Answer:

General equation of circle is :

x²+y²+2 g x + 2 f y +c=0

Having center ,(-g,-f) and radius $=\sqrt{g^2+f^2-c}$

Since the circle passes through (6,6) , (3,-7) and (3,3).

it will satisfy the general equation of circle.

1.→→6²+6²+12 g + 12 f +c=0

72+12 g + 12 f +c=0

2.→ 9 +49 +6 g - 14 f+c=0

6 g -1 4 f + c+58=0

3.→→9+9+6 g + 6 f+c=0

6 g + 6 f + c+18=0

Equation (3)- equation (2)

2 0 f -40=0

20 f = 40

f=2

Equation (1) - 2 equation (2)

40 f-c+72-116=0

40 f -c=44

c=40 ×2-44

c=36

Substituting the value of , f and c in equation (3)

6 g +  6 ×2 +36+18=0

6 g +66=0

6 g=-66

g= -11

Center of circle =(g,f)=(-11,2)