Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If we are given a circle and its equation and a point which lies on it..

Can we find the diametrical opposite point?

share|cite|improve this question
Yes, this is just symmetry. – Karolis Juodelė Nov 14 '13 at 17:38

Say the center of your circle is the point $C=(c_x,x_y)$ in the plane. Translate also the point $P$ you are given by $C$, than the diametrical opposite of the translated point $P-C$ is just its opposite $-P+C$. Translate this back and you have your answer: $$ P' = 2C-P$$

share|cite|improve this answer

Yes, you have the equation $(x-a)^2+(y-b)^2=r^2$. So this is your circle with radius $r$ and center $(a,b)$. Given a point $(m,n)$ on the circle, just create a line connecting the center of the circle $(a,b)$ to $(m,n)$. This new line will intersect the circle at two points, one of which is $(m,n)$ and the other will be the diametrically opposite point on the circle to $(m,n)$. You need only solve for the other intersection point.

share|cite|improve this answer

Sure. You know the center. Hence, you can write the equation that represents the line drawn between the center and the point you are given.

Then solve the two system of equations: the one for the circle and the one for the line segment. The solutions will represent the original point and the one opposite to it.

share|cite|improve this answer

Assume that your circle has the equation $(x-p)^2+(y-q)^2=r^2$, where $(p,q)$ is the centre and $r$ is the radius. Let $(a,b)$ be a point on the circle.

To vector joining the centre $(p,q)$ to the point $(a,b)$ is $(a-p,b-q)$. This is because $$(p,q)+(a-p,b-q)=(a,b)$$

To get to the point diametrically opposed to $(a,b)$ we have to do the opposite of what we did to get from the centre to $(a,b)$. That means, we must do:

$$(p,q) - (a-p,b-q) = (2p-a,2q-b)$$

share|cite|improve this answer

Hint: Find center point (xc,yc) and radius. If given point is (a,b) and required pint ( x,y), then, x + a = 2 xc and y + b = 2 yc as for center points, so find (x,y).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.