Question

The general equation for a circle is a(x2+y2)+bx+cy+d=0. There is exactly one circle passing through the points (1,2),(3,−1), and (0,0). Find an equation for this circle.

Answer 1

Answer:

Step-by-step explanation:

let the eq. of circle be a(x²+y²)+bx+cy+d=0

x²+y²+b/a x+c/a y+d/a=0

if it passes through (0,0)

0²+0²+b/a*0+c/a*0+d/a=0

so d=0

Hence x²+y²+b/a x+c/a y=0

if it passes through (1,2) and (3,-1)

1²+2²+b/a×1+c/a×2=0

5+b/a+2c/a=0   ...(1)

and 3²+(-1)²+b/a*3+c/a*(-1)=0

10+3b/a-c/a=0   ...(2)

(1)+2(2) gives

5+b/a+2 c/a+20+6 b/a -2 c/a=0

7 b/a+25=0

b/a=-25/7

from (1)

5-25/7+2 c/a=0

2 c/a+10/7=0

2 c/a=-10/7

c/a=-5/7

so eq. of circle is x²+y²-25/7 x-5/7 y=0

7(x²+y²)-25 x-5 y=0