Question

Find the equation of circle(s) at any point on which the line joining the points (-1,1) and (5,5) subtends an angle of ${45}^{\circ }$.

Answer 1

Any circle through two points A, B by (n), is
$(x+1)(x+5)(y-1)(y-5)+\lambda (2x-3y+5)=0$
or $x^2+y^2-4x-6y+\lambda (2x-3y+5)=0$......(1)
or $x^2+y^2-x(2\lambda -4)-y(-3\lambda -6)+5\lambda =0$
Since AB students ${45}^{\circ }$ at any point P , hence angle subtended by AB at centre C is ${90}^{\circ }$
Clearly CM is right bisector of AB
$\therefore AM=CM=BM=\frac{1}{2}AB=\sqrt{13}$
$\therefore r=AC=\sqrt{13+13}=\sqrt{26}$
$\therefore 26=r^2=g^2+f^2-c$
${\left(\frac{2\lambda -4}{2}\right)}^2+{\left(\frac{-3\lambda -6}{2}\right)}^2-5\lambda =26$
or $4({\lambda }^2-4\lambda +4)+9({\lambda }^2-4\lambda +4)-20\lambda =104$
or $13{\lambda }^2-52=0\lambda =\pm 2$
Putting the values of $\lambda$ in (1) , the required circles
are $x^2+y^2-12y+10=0$
and $x^2+y^2-8x-10=0$