Finding the equation of a circle.

Question

A circle of radius $2$ lies in the first quadrant touching both axis. Find the equation of the circle centered at $(6,5)$ and touching the above circle externally.

Let me share how I answered this question with your suggestions. Since the radius of the first circle is $2$ and touches both axes. It follows that $h,k$ is $(2,2)$. The formula for this circle is

$$(x-2)^2 + (y-2)^2 = 4 .$$

Given that the center of the second circle is $(6,5)$, then we can get the distance to another circle. The distance between $(6,5)$ and $(2,2)$ is $5$.

We can subtract and get the radius of the second circle. $r=5-2$. Then, the radius is $3$.

The equation of the second circle is

$$(x-6)^2 + (y-5)^5 = 9 .$$

Thank you everyone! :)

Answer 1

The first circle is centered at $(2,2)$, so the distance between two centers of both the circles would be $\sqrt{(6-2)^2 +(5-2)^2} =5$. Hence the radius of circle centred at(6,5) is 3 units. So the final equation of circle is $(x-6)^2+(y-5)^2 = 9$

Answer 2

Equation of circle of radius 2 and touching the axes is $(x-2)^2+(y-2)^2=4$ and its center is $(2,2)$.

Center of required circle is (6,5).Radius of required circle is half of distance between (6,5) and (2,2) because these two circles touch externally.Therefore radius of required circle$=r=5-2$

Now use $(x-h)^2+(y-k)^2=r^2$ and get the equation.