Find condition on a and b so that two circles meet at exactly one point

Question

I have a question that states:

Find a condition on $a$ and $b$ so that the curve $x^2+y^2=1$ touches the curve $(x-a)^2+(y-b)^2=r^2$ at exactly one point.

I know that two circles meet at exactly one point if the distance between the two radii is equal to the radius of the first circle plus the radius of the second circle.

I also know that the radius of the first circle is $1$ and the centre of the first circle is at the origin $(0,0)$.

How would I go about solving the question?

Answer 1

The equation $x^2+y^2 = 1$ represents a circle whose center is at $(0,0)$ and radius is $1$ unit , while $(x-a)^2+(y-b)^2 = r^2$ represents a circle whose center is at $(a,b)$ and radius is $r$ unit.

For the circles to meet at one point , the distance between their radius should be the sum \ difference of their radii .

Distance between $(0,0)$ and $(a,b)$ is $\sqrt{a^2+b^2}$

Radius of unit circle : $1$

Radius of second circle : $r$

So , the required condition is : $\boxed{\sqrt{a^2+b^2} = r\pm1}$

Answer 2

Solving $x^2+y^2=1$ and $(x-a)^2+(y-b)^2=r^2$ as a system of equations, we get that $$-{a}^{2} \left( {a}^{2}+{b}^{2}-{r}^{2}-2\,r-1 \right) \left( {a}^{2} +{b}^{2}-{r}^{2}+2\,r-1 \right) $$ must be zero.