Question

The circle cutting $x^2+y^2-6x+4y-12=0$ orthogonally and having centre $(-1,2)$ is

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

Answer 1

The correct option is A $x^2+y^2+2x-4y-2=0$

If two circle having radii $r_{1}and{r}_2$ and centre at $c_{1}and{c}_2$ cutting each other orthogonally then,
$\cos (180-{90}^{\circ })=\frac{r_1^2+r_2^2-(c_1{c}_2{)}^2}{2r_1{r}_2}=\cos {90}^{\circ }$
$\therefore r_1^2+r_2^2=(c_1{c}_2{)}^2$
Now, $x^2+y^2-6x+4y-12=0$
$r_1=\sqrt{3^2+2^2+12}=5,c_1\equiv (3,-2)$
And other circle has centre $c_2\equiv (-1,2)$
$\therefore r_1^2+r_2^2=c_1{c_2}^2$
$\Rightarrow (5{)}^2+{r_2}^2=(4\sqrt{2}{)}^2$
$\Rightarrow {r_2}^2=32-25$
$\Rightarrow {r_2}^2=7$
$\Rightarrow r_2=\sqrt{7}$
So, equation of circle having centre at $c_2\equiv (-1,2)$ and radius $=\sqrt{7}$ is
$(x+1{)}^2+(y-2{)}^2=r_2^2=7$
$\Rightarrow x^2+y^2+2x-4y-2=0$