Question

The general equation for a circle is

a(x2+y2)+bx+cy+d=0.

a(x2+y2)+bx+cy+d=0.

There is exactly one circle passing through the points (1,1),(−1,−2),(1,1),(−1,−2), and (0,0).(0,0). Find an equation for this circle.

Answer 1

Let the equation of the required circle be

x^2+y^2+bx+cy+d=0 ……………….(i)

According to the problem, the above equation passes through the coordinate points (1,1),(−1,−2),and (0,0) .

Therefore, substituting the coordinates of three points (1,1),(−1,−2),and (0,0) successively in equation (i) we get,

For the point (1, 1):

1^2+1^2+b*1+c*1+d=0       

⇒ b+c+d = -2 ……………….(ii)

For the point (−1,−2):

(-1)^2+(-2)^2+b*(-1)+c*(-2)+d=0       

⇒ -b-2c+d= -5 ……………….(iii)

For the point (0,0):

(0)^2+(0)^2+b*(0)+c*(0)+d=0       

⇒d=0 ……………….(iv)

Substitute d =0 in equation .(ii) and .(iii) to obtain

b + c = -2

-b-2c= -5

Add both

-c= -7

c=7

hence

b= -7-2 = -9

Thus, equation is

x^2+y^2 -9x+7y=0