The point of contact of between two circles and common tangent at this point.

Question

A large circle and a small circle have equations $x^2+y^2+2x-4y-27=0 $ and $x^2+y^2-12x+10y+43=0$ respectively.

a) Show that the two circles externally touch at a single point and find the point of contact.

b) Establish the equation of the common tangent at this point.

---

Answers:

a) Proof and $(3,-2)$

b) $y=x-5$

I need explanation provided please.

Answer 1

a) Solve the system $$x^2+y^2+2x-4y-27=0$$ $$x^2+y^2-12x+10y+43=0$$ There should be only one solution, which is $(3,-2)$.

b) Simple: take the derivative of one of the functions at $x=3$.

Answer 2

$\mathcal{C_1}: (x+1)^2 + (y-2)^2 = (4\sqrt{2})^2\Rightarrow C_1 = (-1,2), r_1 = 4\sqrt{2}$.

$\mathcal{C_2}: (x-6)^2 + (y+5)^2 = (3\sqrt{2})^2\Rightarrow C_2 = (6,-5), r_2 = 3\sqrt{2}$.

$\overline{C_1C_2} = \sqrt{7^2+7^2} = 7\sqrt{2} = r_1+r_2\Rightarrow \mathcal{C_1}, \mathcal{C_2} $ are touching each other as claimed.

Answer 3

The larger circle is: $$(x+1)^2+(y-2)^2=32$$ So its center is $(-1,2)$ and radius is $\sqrt{32}$

The smaller circle is: $$(x-6)^2+(y+5)^2=8$$ So its center is $(6,-5)$ and radius is $\sqrt{8}$.

Compute the distance between centers and compare it with the sum of the radii to check tangency.

For getting the tangent line (without derivatives, since you have labeled it as pre-calculus).

In general the intersection of two circles (and its geometry) can be viewed by using the following: $$x^2+y^2+2x-4y-27 + \lambda(x^2+y^2-12x+10y+43)=0$$ where $\lambda$ is a parameter which will help you figure out all possible scenarios. The equation stated above can be re-written as: $$(1+\lambda)x^2+(1+\lambda)y^2+(2-12\lambda)x+(10\lambda-4)y+(43\lambda-27)=0$$

So if you want a line that passes through the intersection of these two circles then you want the coefficient of the second degree terms to be zero (``linear'').

Thus you want $\lambda=-1$. This gives you

$$14x-14y=70$$ which is same as $$x-y=5.$$

Answer 4

The equation of family of circles touching a given circle 'S' and a line 'L' is given by: \begin{equation} S+ \lambda L=0. \end{equation}

If we consider one of the circles then,

\begin{equation} (x^2+y^2+2x-4y-27)+\lambda(y-mx-c)=0. \end{equation}

This must give the equation of the other circle making contact with $x^2+y^2+2x-4y-27=0$ and, y-mx-c=0 is the equation of the common tangent. Thus we can write,

\begin{equation} (x^2+y^2+2x-4y-27)+\lambda(y-mx-c)=x^2+y^2-12x+10y+43. \end{equation}

By comparing the coefficients of 'x', 'y' and the constant we can get the following values,

\begin{equation} \lambda=14, \quad c=-5, \quad m=1. \end{equation}

Hence, the equation of tangent becomes, y=x-5. Next, to find the point of contact we can find the equation of normal at this point of contact. As this normal would pass through the centre of each circle, we can use their centres such as (-1,2) and (6,-5) to get the equation of normal as,

\begin{equation} y=1-x. \end{equation}

Finding the point of intersection of the normal and the common tangent would give us the point of contact as (3,-2).