Question

The equation of the image of the circle $x^2+y^2-6x-4y=0$ in the bisector of $2^{\text{nd}}$ and $4^{\text{th}}$ quadrant is

• A. $x^2+y^2-4x-6y=0$
• B. $x^2+y^2+4x-6y=0$
• C. $x^2+y^2-4x+6y=0$
• D. $x^2+y^2+4x+6y=0$

Answer 1

Answer: (D)

$x^2+y^2+4x+6y=0$

Given circle is $x^2+y^2-6x-4y=0$
Centre is $C=(3,2)$
Radius is $r=\sqrt{9+4-0}=\sqrt{13}$
The bisector of $2^{\text{nd}}$ and $4^{\text{th}}$ quadrant is
$x+y=0$

Let new center will be $C^{\prime }(h,k)$, then
Now, the image if the centre is
$\frac{h-3}{1}=\frac{k-2}{1}=-\frac{2(3+2)}{2}\Rightarrow (h,k)=(-2,-3)$

Hence, the required equation of the circle is
$(x+2{)}^2+(y+3{)}^2=13\therefore x^2+y^2+4x+6y=0$