Question

The equation of the image of circle $x^2+y^2-4x+6y+10=0$ in the line $4x-3y+8=0$ is

• A. $(x-6{)}^2+(y-3{)}^2=3$
• B. $(x+6{)}^2+(y-3{)}^2=3$
• C. $(x+6{)}^2+(y+3{)}^2=3$
• D. $(x-6{)}^2+(y+3{)}^2=3$

Answer 1

The correct option is B $(x+6{)}^2+(y-3{)}^2=3$

Given circle is, $x^2+y^2-4x+6y+10=0$
Comparing with general circle $x^2+y^2+2gx+2fy+c=0,$ we get
$g=-2,f=3,c=10$
The radius and centre of the circle is
radius, $r=\sqrt{g^2+f^2-c}$
$\Rightarrow r=\sqrt{(-2{)}^2+3^2-10}=\sqrt{3}C\equiv (-g,-f)\equiv (2,-3)$

Now, the image of the centre in the line is
$\frac{x-2}{4}=\frac{y+3}{-3}=-\frac{2(4\times 2-3\times (-3)+8)}{4^2+(-3{)}^2}$
Solving above equations, we get
$x=-6,y=3$
So, the centre of image circle is $C_1=(-6,3)$
And radius is same as $\sqrt{3}.$

$\therefore$ Equation of image circle is
${(x+6)}^2+{(y-3)}^2=3$