Question

Find the equation of the circle which pass through the points A (-1 , 4) and B (1,2) and which touches the line $3x-y-3=0$

Answer 1

The equation of circle through A and B as in part (a) is $S+\lambda P=0$
or $(x+1)(x-1)+(y-4)(y-2)+\lambda (x+y-3)=0$
or $x^2+y^2+\lambda x+(\lambda -6)y+(7-3\lambda )=0$
The circle touches the line 3x - y - 3 = 0.We may apply the condition of tangency i.e.p = r and find two values of $\lambda$ yourself.
Another method is that if the line is a tangent then it will cut the circle in two coincident points.
Putting y = 3(x -1 ) in the equation of circle, we get
$x^2+9(x-1{)}^2+\lambda x+3(x-1\left)\right(\lambda -6)+7-3\lambda =0$
or $10x^2+(-36+4\lambda )x+(34-6\lambda )=0$
The roots of the above quadratic should be equal
$\therefore \Delta =0$
${\lambda }^2-18\lambda +81-5(17-3\lambda )=0$
${\lambda }^2-3\lambda -4=0or\lambda =4,-1$
Hence the circles are :
$x^2+y^2+4x-2y-5=0$
and $x^2+y^2-x-7y+10=0$