Problem Involving Circle Geometry

Question

A circle passes through the following points: $(0,0)$ $(1,3)$ $(3,0)$ and $(2,3)$

Find the centre and radius of the circle and explain why. Thanks for any help. So far I have drawn a graph, the centre is around (1.5,1.2) and the radius around 1.9. But I need The surds and haven't been able to find the exact points.

Answer 1

Let us begin by drawing a sketch. Even a rough one will notify you that if the circle exists, it must have $x-$coordinate $\frac{3}{2}$ because that is the $x-$coordinate for the center of the points.

We now apply distance formula. With the points $(3, 0)$ and $(2, 3),$ and our knowledge that the $x-$coordinate of the center is $\frac{3}{2},$ we can write $$\sqrt{\frac{9}{4} + y^{2}} = \sqrt{\frac{1}{4} + (y - 3)^{2}}$$ $$2 + y^{2} = (y - 3)^{2}$$ $$2 = -6y + 9$$ $$y = \frac{7}{6}.$$

The center is thus $\boxed{(\frac{3}{2}, \frac{7}{6})}.$

We can easily find the radius as follows: $$r = \sqrt{\frac{9}{4} + \frac{49}{36}}$$ $$= \sqrt{\frac{85}{36}}$$ $$= \boxed{\frac{\sqrt{130}}{6}}.$$

Answer 2

According to google the definition of a circle is:

A round plane figure whose boundary (the circumference) consists of points equidistant from a fixed point (the center).

Mathematically we can say that any points that lie on this curve must satisfy the equation:

$$(x-x_0)^2+(y-y_0)^2=r^2$$

where $(x_0,y_0)$ is the center and r is the radius. This means that all of your points P1(0,0), P2(1,3), P3(3,0), and P4(2,3) must satisfy this condition. Using these points provides you with 4 equations and 3 unknowns. Although the system is $over-constrained$ it is solvable and can get you the center and radius.

Starting off let us look at P1 and P2, respectively they yield the following. $$(x_1-x_0)^2+(y_1-y_0)^2=r^2 \rightarrow (0-x_0)^2+(0-y_0)^2=r^2$$ $$(x_2-x_0)^2+(y_2-y_0)^2=r^2 \rightarrow (1-x_0)^2+(3-y_0)^2=r^2$$ If we subtract the first equation from the second equation we get: $$(1-x_0)^2+(3-y_0)^2-((0-x_0)^2+(0-y_0)^2)=r^2-r^2$$ $$(1-x_0)^2+(3-y_0)^2-x_0^2-y_0^2=0 \rightarrow y_0= -\frac{2}{3}x_0+\frac{10}{3}$$

Now that we have $y_0$ in terms of $x_0$ I think it would be a good place for you to pick up and solve for $x_0$ and r.

Good Luck.

Answer 3

Here's how to do it.

1) Draw the trapezoid containing the given points as vertices.

2) Identify with equations the perendicular bisectors of all the sides of the trapezoid. You should be able to see that the two bases have the same perpendicular bisector, a necessary condition for tour trapezoid to be inscribed in a circle.

3) Any two distinct perpendicular bisectors will intersect at the center of the circle. Usually you would have to solve simultaneously for $x$ and $y$, but here the perpendicular bisector of both bases will allow you to get $x$ separately and then you can calculate $y$.

4) Once you have the center get the radius to any point in the circle with the usual distance formula.

5) Post your own answer here, get upvotes and win a self-learber badge. (OK, that part is not mathematics, but is is true on SE.)