Question

If a circle of radius 3 units rolls inside of the circle $(x-2{)}^2+(y-1{)}^2=25$, then the locus of the centre of that circle is a circle of diameter

• A. $2$
• B. $8$
• C. $2\sqrt{2}$
• D. $4$

Answer 1

Answer: (D)

$4$

$(x-2{)}^2+{(y-1)}^2={\left(5\right)}^2$
$C_1=(2,1)$ and $r_1=5$
Let $C(x,y)$ be the centre of the circle having radius $3$
Distance between the centres is
$\therefore {(x-2)}^2+{(y-1)}^2=4$
$x^2+y^2-4x-2y+1=0$
This is the locus of centre of the circle of radius $3$
$\therefore r=\sqrt{(2{)}^2+(1{)}^2-1}=2$
So, diameter of locus circle is $=4$