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I have these two circles, $\left(-6,\ 0.8\right)$ for the blue circle, and $\left(-3.988,\ -8.159\right)$ for the black circle.

Using the formula to find the mid-point of these circles, I got $\left(-4.994,\ -3.6795\right)$.

For my assignment I would have to find the reflection line of these two shapes but I don't know how, even after searching it up on the internet. I could only find tutorials to find the reflected shape after drawing a reflection line.

If someone knows how to help me with this that would be great.

(this is my first post meaning I'm not good at using forums like this, so please don't mind if this post is not the best)

Minh Tri Nguyen Le
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  • Let $C$ and $C'$ be these points (centers of circles) ; you have first to determine the midpoint $M$ of line segment [CC']. Do you know how ? – Jean Marie Nov 16 '22 at 22:38
  • Determine now the equation $y=mx+p$ of line CC'. Then you have to define the perpendicular line $y=m'x+p'$ passing through $M$ to line segment [CC']. Do you know that ortogonality mens that the product of slopes $mm'=-1$ ? Now find p' by the fact that $M$ belongs to this line. – Jean Marie Nov 16 '22 at 22:44
  • @JeanMarie I used the formula $\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)$ and got $\left(-4.994,\ -3.6795\right)$. Is that what you mean? – Minh Tri Nguyen Le Nov 16 '22 at 22:44
  • Yes. Now proceed with the method I have indicated. – Jean Marie Nov 16 '22 at 22:45
  • @JeanMarie Thanks so much. It worked – Minh Tri Nguyen Le Nov 17 '22 at 06:39

1 Answers1

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You already found Center Point coordinates $P (xC,yC)$

Let $(x1,y1)=\left(-6,\ 0.8\right)$ and $(x2,y2)=\left(-3.988,\ -8.159\right)$

Slope of centerline is

$$ m=\frac{y2-y1}{x2-x1}$$

Slope of any perpendicular line to the center line is $$-\frac{1}{m} $$

Reflection Line equation

$$ \frac{y-yC}{x-xC} = -\frac{1}{m}$$

Narasimham
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