Question

The equation of the image of circle $x^2+y^2-2x-4y=0$ in the line $3x-4y+30=0$ is

• A. $(x-5{)}^2+(y-10{)}^2=5$
• B. $(x+5{)}^2+(y+10{)}^2=5$
• C. $(x+5{)}^2+(y-10{)}^2=5$
• D. $(x-5{)}^2+(y+10{)}^2=5$

Answer 1

The correct option is C $(x+5{)}^2+(y-10{)}^2=5$

Given circle is $x^2+y^2-2x-4y=0$
which is in the form of $x^2+y^2+2gx+2fy+c=0.$
Then $g=-1,f=-2,c=0$
The centre and radius of the circle is
$C=(1,2)$
and $r=\sqrt{g^2+f^2-c}=\sqrt{1+4-0}$
$\Rightarrow r=\sqrt{5}$

Now, the image of the centre in the given line is
$\frac{x-1}{3}=\frac{y-2}{-4}=-\frac{2(3\times 1-4\times 2+30)}{3^2+(-4{)}^2}\Rightarrow x=-5,y=10$

So, the centre of image circle is
$C_1=(-5,10)$
And radius is same as $\sqrt{5}$

Hence, the equation of image circle is
$(x+5{)}^2+(y-10{)}^2=5$