Question

The equation of the image of the circle $(x-2{)}^2+(y-2{)}^2=1$ by the mirror $x+y=19$ is

• A. $(x-14{)}^2+(y-13{)}^2=1$
• B. $(x-15{)}^2+(y-14{)}^2=1$
• C. $(x-16{)}^2+(y-15{)}^2=1$
• D. $(x-17{)}^2+(y-16{)}^2=1$

Answer 1

The correct option is D $(x-17{)}^2+(y-16{)}^2=1$

Consider the given equation of the circle

$(x-3{)}^2+(y-2{)}^2=1$ ……..$\left(1\right)$

comparing that

$(x-h{)}^2+(y-k{)}^2=a^2$

$(h,k)=(3,2)$ and $a=1$

equation of mirror is

$x+y=19$

$x+y-19=0$ …….$\left(2\right)$

Let $(\alpha ,\beta )$ be the centre of required circle. Then,

We know that,

$\frac{\alpha -h}{a}=\frac{\beta -k}{a}=\frac{-2(x+y-19)}{a^2+a^2}$

$\frac{\alpha -3}{a}=\frac{\beta -2}{a}=\frac{-2(3+2-19)}{2a^2}$

$\frac{\alpha -3}{1}=\frac{\beta -2}{1}=\frac{-2(5-19)}{2\times 1^2}$

$\alpha -3=\beta -2=\frac{28}{2}$

$\alpha -3=\beta -2=14$

Now,

$\alpha -3=14$, $\beta -2=14$

$\alpha =14+3=17$, $\beta =14+2$

$\alpha =17$, $\beta =16$

Hence, the required circle is

$(x-\alpha {)}^2+(y-\beta {)}^2=a^2$

$(x-17{)}^2+(y-16{)}^2=1^2$

$(x-17{)}^2+(y-16{)}^2=1$

Hence, this is the answer.