Question

A circle $C$ passes through the points $A(2,3)$ & $B(-1,1)$, and its centre lies on the line $L:x-3y-11=0$. If the origin is shifted to the point $(3,-1)$ after translation of axes, then find the new coordinates of the point $A$ & $B$, and also the new equation of the line $L$. Hence find the new equation of circle $C$.

Answer 1

Let $(x_1,y_1)$ be center of circle

$\Rightarrow x_1-3y_1-11=0\to 1$

Also, ${(x_1-2)}^2+{(y_1-3)}^2={(x_1+1)}^2+{(y_1-1)}^2$

$\Rightarrow x_1^2-4x_1+4+y_1^2-6y_1+9=x_1^2+2x_1+1+y_1^2-2y_1+1$

$\Rightarrow 6x_1+4y_1-11=0\to 2$

Solving $1$ and $2$, we get

Center$=(-\frac{7}{2},\frac{5}{2})$

Radius$=\frac{\sqrt{122}}{2}$

Now, new co ordinates of $A=(2-3,3-1)$

$=(-1,2)$

$B=((-1-3),1-1)$

$=(-4,0)$

New equation of line$=y=\frac{x-3}{3}-\frac{11}{3}+1$

$\Rightarrow 3y=x-11$

New center$=(\frac{-7}{2}-3,\frac{5}{2}-1)$

$=(\frac{-13}{2},\frac{3}{2})$

$\therefore$New equation of circle$={(x+\frac{13}{2})}^2+{(y-\frac{3}{2})}^2=\frac{122}{4}$