Review question

# Can we show that these two circles touch? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6100

## Solution

Show that the circles having equations $x^2 + y^2 = 25$ and $x^2 + y^2 - 24x - 18y + 125 = 0$ touch each other. Calculate the coordinates of the point at which they touch.

#### Approach 1: Geometric

The circle with equation $x^2+y^2=25$ has its centre at the origin and radius $5$.

We can complete the square for the equation $x^2 + y^2 - 24x - 18y + 125 = 0$ to find its centre and radius: $\begin{gathered} x^2 + y^2 - 24x - 18y + 125 = 0\\ \implies\quad (x-12)^2-12^2 + (y-9)^2 -9^2 + 125 = 0\\ \implies\quad (x-12)^2 + (y-9)^2 = 144+81-125\\ \implies\quad (x-12)^2 + (y-9)^2 = 100 \end{gathered}$ so this circle has centre $(12,9)$ and radius $10$.

We can sketch these:

It seems as though the circles touch. Let’s prove that!

The length of the line joining the two centres $(0,0)$ and $(12,9)$ is (using Pythagoras) $\sqrt{12^2+9^2}=\sqrt{144+81}=\sqrt{225}=15.$

Alternatively, we can notice that $9$ and $12$ form the two short sides of right-angled triangle, which must be a $3$-$4$-$5$ triangle scaled by a factor of $3$, and therefore the diagonal length is $3\times 5=15$.

Now the radii of the two circles are $5$ and $10$. Since $5+10=15$ (the distance between the centres), the two circles touch.

To find the coordinates of the point where they touch, we can use similar triangles:

The small triangle has sides in the ratio $a:b:5$ (base to height to hypotenuse), while in the large triangle, they are in the ratio $12:9:15$. These ratios are the same, so $a:b:5=4:3:5$, and the point of touching has coordinates $(4,3)$.

Another way to think about this is that the point of touching is $\dfrac{5}{15}=\dfrac{1}{3}$ of the way along the line from $(0,0)$ to $(12,9)$, so it is at $(4,3)$.

#### Approach 2: Algebraic

For the circles to touch, we must show that they intersect at exactly one point.

We have the equations of the two circles: $x^2 + y^2 = 25$ and $x^2 + y^2 - 24x - 18y + 125 = 0.$

Substituting the first equation into the second (or alternatively subtracting the first equation from the second) gives \begin{align*} &25 - 24x - 18y + 125 = 0\\ \Longrightarrow\quad& 24x + 18y = 150\\ \Longrightarrow\quad& 4x + 3y = 25. \end{align*}

So if the two circles meet in two points, they must both lie on $4x + 3y = 25.$

We can rearrange $4x + 3y = 25$ to get $x = \frac{1}{4}(25-3y)$ and substitute into the equation $x^2 + y^2 = 25$: \begin{align*} &\left(\frac{1}{4}(25-3y)\right)^2 +y^2 = 25\\ \iff\quad& 25y^2 - 150y + 225 = 0\\ \iff\quad& y^2 - 6y + 9 = 0\\ \iff\quad& (y-3)^2 = 0\\ \iff\quad& y = 3. \end{align*}

The quadratic equation has a repeated root, so has exactly one solution.

This means that the line $4x + 3y = 25$ is a tangent to the circle $x^2 + y^2 = 25.$

We can find the corresponding $x$ value by setting $y = 3$ in the equation $4x + 3y = 25$: \begin{align*} &4x + 3(3) = 25\\ \Longrightarrow\quad& x = 4. \end{align*} Substituting $x = \frac{1}{4}(25-3y)$ into the equation $x^2 + y^2 - 24x - 18y + 125 = 0,$ we have \begin{align*} &\left(\frac{1}{4}(25-3y)\right)^2 +y^2 -24\left(\frac{1}{4}(25-3y)\right)-18y+125= 0\\ \iff\quad& 25y^2 - 150y + 225 = 0\\ \iff\quad& y^2 - 6y + 9 = 0\\ \iff\quad& (y-3)^2 = 0\\ \iff\quad& y = 3. \end{align*}

This means that the line $4x + 3y = 25$ is a tangent to the second circle also at $(4,3)$.

So the circles $x^2 + y^2 = 25$ and $x^2 + y^2 - 24x - 18y + 125 = 0$ touch at the point $(4,3)$.