The image of the circle $(x - 3)^2 + (y - 2)^2 = 1$ in the line mirror $ax + by = 19$ is $(x - 1)^2 + (y - 16)^2 = 1$ then values of (a, b) is

The eqn. of the image of the circle $(x-3)^{2}+(y-2)^{2}=1$ in the mirror $x+y=19$ is

$\begin{array}{c} c(3,2), r=1 \\ \gamma^{\prime}=1 . \end{array}$

$\Rightarrow$ formula $\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{-2\left(a x_{1}+b y_{1}+c\right)}{a^{2}+b^{2}}$

$a=1, b=1, c=-15 .$

$\begin{array}{c} \text { mirror image } \Rightarrow \frac{x-3}{1}=\frac{y-2}{1}=\frac{-2(3+2-19)}{2} \\ x-3=y-2=1 y \\ \therefore x=17, y=16 \end{array}$

$\begin{array}{l} c^{\prime}(17,16) \quad, r=1 \\ \Rightarrow(x-17)^{2}+(4-16)^{2}=1 \\ \therefore\mid(x-17)^{2}+(y-16)^{2}=1 \end{array}$