Question

If the power of point $(1,-2)$ with respect to $x^2+y^2=1$ is equal to the radius of a circle and $(3,2)$ is the centre of that circle, then the equation of that circle is

• A. $x^2+y^2-6x-4y-3=0$
• B. $x^2+y^2-6x-4y-5=0$
• C. $x^2+y^2-6x-4y+3=0$
• D. $x^2+y^2-6x-4y+5=0$

Answer 1

The correct option is A $x^2+y^2-6x-4y-3=0$

The power of the point $(1,-2)$ with respect to the circle $x^2+y^2=1$ is
$S_1=(1{)}^2+(-2{)}^2-1=4$
$\therefore$ radius, $r=4$
Then the equation of circle whose centre $C\equiv (3,2)$ and radius $r=4$ is
$(x-3{)}^2+(y-2{)}^2=4^2$
$\Rightarrow x^2+y^2-6x-4y-3=0$