Question

A circle of radius $2$ unit lies in the first quadrant and touches both the axes of coordinates, equation of circle with center at $(6,5)$ and touching the above circle externally is

• A. $(x-6{)}^2+(y-5{)}^2=9$
• B. $x^2+y^2-10x-12y+52=0$
• C. $x^2+y^2-12x-10y+32=0$
• D. $x^2+y^2+12x-10y-32=0$

Answer 1

The correct option is A $(x-6{)}^2+(y-5{)}^2=9$

Let the center of the given circle be $(h,k)$.

The circle of radius 2 lies in the first quadrant and touches both the axes.

$\therefore h=k=2$

The equation of the given circle is

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

$\Rightarrow (x-1{)}^2+(y-2{)}^2=4$-----------------(1)

Let the radius of the required circle be $r$.

$\therefore$ The equation of the required circle is

$(x-6{)}^2+(y-5{)}^2=r^2$-------------------------(2)

Given, Circle 1 & Circle 2 touch each other externally.

$\therefore$ Sum of their radius$=$Distance between their centers.

$\Rightarrow r+2=\sqrt{(6-2{)}^2+(5-2{)}^2}$

$\Rightarrow r+2=\sqrt{16+9}=5$

$\therefore r=3$

Hence, the equation of required circle is

$(x-6{)}^2+(y-5{)}^2=3^2$---------------------(Using Equation (2))

$\Rightarrow (x-6{)}^2+(y-5{)}^2=9$