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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?

The Beast
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2 Answers2

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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 : $\boxed{\sqrt{a^2+b^2} = r\pm1}$

The Demonix _ Hermit
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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.

Dr. Sonnhard Graubner
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  • Perhaps you can explain how you got that result (which looks like copy/paste from some CAS system)? – Martin R Dec 14 '19 at 18:02
  • Write your second equation as $$x^2+y^2-2ax-2by+a^2+b^2=r^2$$ and substitute $$x^2+y^2=1$$ this means you have $$1-2ax-2by+a^2+b^2=r^2$$ a linear e2quation in $x,y$ – Dr. Sonnhard Graubner Dec 14 '19 at 18:06