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The two circles $x^2 + y^2−16 x−20 y + 115 =0$ and $x^2 + y^2+8 x−10 y + 5 =0$ are tangent. How could I find the point of tangency?

KReiser
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Anas Yasser
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  • Can you solve the system of equations to find the point that lies on both? What have you tried? – Calvin Lin Jun 14 '20 at 23:33
  • Start by completing the square for both equations, then solve the system of equations. – dude076 Jun 14 '20 at 23:37
  • The difference of the 2 equations gives you the tangent line (Do you see why?). Taking that to intersect one of the circles given you the point of tangency. Alternatively, take the tangent line intersect the line of centers to get the point of tangency. – Calvin Lin Jun 14 '20 at 23:37
  • thank you very much – Anas Yasser Jun 14 '20 at 23:40

1 Answers1

1

$x^2+y^2-16x-20y+115=x^2+y^2+8x-10y+5$

$\Longleftrightarrow -16x-20y+110=8x-10y$

$\Longleftrightarrow -24x-10y+110=0$

$-24x-10y+110=0$ is the tangent to both circles.

The tangent point of the two circles is where the line between the two center point of the circles and the tangent above cross each other.

$x^2+y^2-16x-20y+115=(x-8)^2+(y-10)^2-49\Longrightarrow C_1=(8|10)$

$x^2+y^2+8x-10y+5=(x+4)^2+(y-5)^2-34 \Longrightarrow C_2=(-4|5)$

Now you only need to connect $C_1$ and $C_2$ to a line and see where it intersects with the tangent we found above!

Are you able to continue from here? :)

Okey we continue:

We found out with rewriting our circle equations where our Center $C_1$ and $C_2$ lie on the plane.

Now we first "draw" a vector between thoose two points.

$\begin{pmatrix} 8 \\ 10 \end{pmatrix}-\begin{pmatrix} -4 \\ 5 \end{pmatrix}=\begin{pmatrix} 12 \\ 5 \end{pmatrix}$ this vector represents our lines growth/slope

and now we "move" the vector to the right position with fixing it at one center point (ofc it will also cross the second ;))

with picking $C_1$ our line will be $h:\begin{pmatrix} 8 \\ 10 \end{pmatrix}+\lambda\begin{pmatrix} 12 \\ 5 \end{pmatrix}$

Now we need to change the line we have into a form which we can calculate with:

$x=8+12\lambda $

$y=10+5 \lambda$

$\Longrightarrow \lambda=\frac{x-8}{12}$

$\Longrightarrow y=10+5\frac{x-8}{12}\Longleftrightarrow y - \frac{5x}{12}-\frac{20}{3}=0$

So your line we were looking for is also called:

$y - \frac{5x}{12}-\frac{20}{3}=0$

Now we only need to set our tangent line equal to this line which is crossing both $C_1$ and $C_2$.

(With $C_1,C_2$ being the center of our circle)

$y - \frac{5x}{12}-\frac{20}{3}=0$

$-24x-10y+110=0$

$\Longrightarrow x= \frac{20}{13},y=\frac{95}{13}$

So finally our tangent point $T=\left(\frac{20}{13}|\frac{95}{13}\right)$

Any more questions? ;)

CoffeeArabica
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