Question

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

• A. $(-1,1)$
• B. $(1,-1)$
• C. $(1,1)$
• D. none of these

Answer 1

The correct option is D none of these

Since, two circles are mirror image of each other w.r.t line $ax+by=19$
Then, 1) mid point of both the center lie on the straight line $ax+by=19$
2) line joining the centers is perpendicular to the given straight line $ax+by=19$
1) mid point is $(2,9)$ and it lies on $ax+by=19$
$\Rightarrow 2a+9b=19$ ....(i)
2) slope of line joining the centers $m_1=\frac{16-2}{1-3}=-7$
slope of line $ax+by=19$ $m_2=\frac{-a}{b}$
$m_1{m}_2=\frac{7a}{b}=-1$ ...(ii)
solving (i) and (ii): $(a,b)=(-\frac{19}{61},\frac{133}{61})$