Consider 3 non collinear points A,B,C with coordinates (0,6),(5,5) and (−1,1) respectively. Equation of a line tangent to the circle circumscribing the triangle ABC and passing through the origin is
Consider 3 non collinear points A,B,C with coordinates (0,6),(5,5) and (−1,1) respectively. Equation of a line tangent to the circle circumscribing the triangle ABC and passing through the origin is
The correct option is C 2x+3y=0
The general equation of a
circle is,
x2+y2+2gx+2fy+c=0
Put (0,6)
in the general equation,
(0)2+(6)2+2g(0)+2f(6)+c=0
36+12f=−c
(1)
Put (5,5)
in the general equation,
(5)2+(5)2+2g(5)+2f(5)+c=0
50+10g+10f=−c
(2)
From equation (1) and (2),
36+12f=50+10g+10f
10g−2f+14=0 (3)
Put (−1,1) in the general equation,
(−1)2+(1)2+2g(−1)+2f(1)+c=0
2−2g+2f=−c
(4)
36+12f=2−2g+2f
2g+10f+34=0 (5)
From equation (3) and (4),
g=−2 and f=−3
Substituting the values in
equation(1),
36+12(−3)=−c
36−36=−c
c=0
Since the general equation
of the circle is,
x2+y2+2gx+2fy+c=0
Let the point (x1,y1) satisfies the equation of the circle, then,
xx1+yy1+2g(x+x1)+2f(y+y1)+c=0
At origin,
x(0)+y(0)+2g(x+0)+2f(y+0)+c=0
2gx+2fy+0=0
gx+fy=0
(−2)x+(−3)y=0
2x+3y=0
The general equation of a circle is,
x2+y2+2gx+2fy+c=0
Put (0,6) in the general equation,
(0)2+(6)2+2g(0)+2f(6)+c=0
36+12f=−c (1)
Put (5,5) in the general equation,
(5)2+(5)2+2g(5)+2f(5)+c=0
50+10g+10f=−c (2)
From equation (1) and (2),
36+12f=50+10g+10f
10g−2f+14=0 (3)
Put (−1,1) in the general equation,
(−1)2+(1)2+2g(−1)+2f(1)+c=0
2−2g+2f=−c (4)
36+12f=2−2g+2f
2g+10f+34=0 (5)
From equation (3) and (4),
g=−2 and f=−3
Substituting the values in equation(1),
36+12(−3)=−c
36−36=−c
c=0
Since the general equation of the circle is,
x2+y2+2gx+2fy+c=0
Let the point (x1,y1) satisfies the equation of the circle, then,
xx1+yy1+2g(x+x1)+2f(y+y1)+c=0
At origin,
x(0)+y(0)+2g(x+0)+2f(y+0)+c=0
2gx+2fy+0=0
gx+fy=0
(−2)x+(−3)y=0
2x+3y=0
