If we are given a circle and its equation and a point which lies on it..can we find the diametrical opposite point?

Question

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

Can we find the diametrical opposite point?

Answer 1

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.

Answer 2

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.

Answer 3

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: $$ P' = 2C-P$$

Answer 4

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:

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