Can we find the equations of two circles that touch another?

Can we find the equations of two circles that touch another?

Question

Show by calculation that the point $P(-9,-3)$ lies inside the circle $x^2+y^2+4x-8y-80=0.$

Find the equations of the two circles which pass through $P$, touch the given circle, and have radii of 5 units.

Suggestion

Can we find the centre and the radius of the circle? How far is $P$ from the centre of the circle?

We know these circles pass through $P$ — can we find another point that these circles must go through?

Solution

Let’s call this circle $D$. Firstly, we rewrite the equation of $D$ in the form $(x-a)^2+(y-b)^2=r^2$, so that we can easily work out its centre and radius.

We have

$$\begin{align*} &x^2+y^2+4x-8y-80=0\\ \iff\quad& (x+2)^2 - 4 + (y-4)^2-16 - 80 =0\\ \iff\quad& (x+2)^2 + (y-4)^2 = 100. \end{align*}$$

$D$ has centre $(-2,4)$ and radius $10$. This means that a point $Q$ lies inside $D$ if and only if the distance between $Q$ and the point $(-2,4)$ is less than $10$.

Let

$d$

be the distance between

$(-9,-3)$

and

$(-2,4)$

. Then

$$\begin{align*} d &= \sqrt{(-9-(-2))^2+(-3-4)^2}\\ &= \sqrt{(-7)^2+(-7)^2}\\ &= \sqrt{49+49}\\ &= \sqrt{98} \end{align*}$$

Now $\sqrt{98} < 10$, so the distance between P and $(-2,4)$ is less than $10$, and $P$ lies inside $D$.

The normal to the circle $C_1$ at $T_1$ is also normal to $D$.

It must go through the centre of $C_1$, but it must go through the centre of $D$ too.

Now $D$ has radius $10$, and $C_1$ has diameter $10$, so we can deduce $C_1$, and similarly $C_2$, goes through $(-2,4)$, the centre of $D$.

Now suppose one of the smaller circles $C$ has centre $(a,b)$.

The circle $C$ passes through $(-9,-3)$, so the distance between $(a,b)$ and $(-9,-3)$ is $5$. So $(a+9)^2 + (b+3)^2 = 25,$ by Pythagoras’ theorem.

But also $C$ passes through $(-2,4)$, so the distance between $(a,b)$ and $(-2,4)$ is $5$. So $(a+2)^2 + (b-4)^2 = 25,$ by Pythagoras’ theorem.

So we have the two equations $a^2+b^2+18a+6b+65=0$ and $a^2+b^2+4a-8b-5=0.$

Subtracting these gives $14a+14b+70=0$, so $70=-14a-14b$ and $b=-5-a.$

Substituting this into $(a+9)^2+(b+3)^2=25$ gives us $a^2+18a+81+4+4a+a^2=25,$ that is, $a^2+11a+30=0.$

This quadratic factorises into $(a+5)(a+6)=0$, so $a = -5$ or $a=-6$.

Substituting these values into $b=-5-a$, we get two possibilities for $(a,b)$, $(-5,0)$ and $(-6,1)$.

It’s easy to check that these points are viable ones for the centre of $C$ (they do satisfy both equations).

So as the circles have radius $5$, we get that the equations for these two circles are $(x+5)^2+y^2=25,$ and $(x+6)^2+(y-1)^2=25.$