Question

Find the equation of director circle of the circle $(x-1{)}^2+(y-2{)}^2=4$

• A. $(x-1{)}^2+(y-2{)}^2=4\sqrt{2}$
• B. $(x-1{)}^2+(y-2{)}^2=8$
• C. $(x+1{)}^2+(y+2{)}^2=4\sqrt{2}$
• D. $(x-2{)}^2+(y-1{)}^2=8$

Answer 1

The correct option is B

$(x-1{)}^2+(y-2{)}^2=8$

Given the circle $(x-1{)}^2+(y-2{)}^2=4$

Center of original circle =(1,2) and radius =$\sqrt{2}$

We know,

Director circle of a circle is the locus of the point of intersection of two perpendicular tangents.

It is the concentric circle having radius equal to $\sqrt{2}$ times the orginal circle.

Center of director circle =(1,2) and radius =$2\sqrt{2}$

Equation of the director circle is $(x-1{)}^2+(y-2{)}^2=(2\sqrt{2}{)}^2$

$(x-1{)}^2+(y-2{)}^2=8$

Option B is correct