Point in area enclosed by a circle?

Question

I need help on how to determine if the point is in the area enclosed by a circle passing through the three vertices of the triangle, given the coordinates of the point and coordinates of the triangles vertices... Help, someone? I need this soon...
Thanks :D

Tnx CrazyCoder, but I'm afraid I have another question, how can I determine the coordinates of the center of the circumcircle? I need a formula...
Sorry for this, but I need it for a programming problem...

Answer 1

Hi,

Because the circumcenter must be the same distance from each of the vertices, and because each side of the triangle is a chord of the circle, it is easy to understand that the circumcircle must be at the intersection of the perpendicular bisectors of the sides of the triangle. To find the circumcenter, find the perpendicular bisectors of any 2 sides of the triangle. Their point of intersection would be the circumcenter. (To verify the answer, find the third perpendicular bisector and verify that it also intersects at the same point.)

The radius of the circumscribing triangle is given by the formula R= a*b*c/(4*area)where a, b, and c are the three side lengths, and the area represents the area of the triangle. If the area is written using Heron's formula, then you can express the radius only in terms of the three side lengths.

Another way to find the radius of the circumcircle is given by the formula:

...............abc
--------------------------------------------- = r
√[(a+b+c)(a+b-c)(b+c-a)(a+c-b)]

where a,b,c are the lengths of the sides of the triangle.


Hopefully, this will give you enough information to find your circle's center and radius so you can determine if the given point is inside of it or not. Good luck on your programming!!

Answer 2

The circumcenter is the point equidistant from the three vertices. It may or may not be inside the triangle itself.

The circumcenter is the intersection of the perpendicular bisectors of the sides. The intersections of any two of the perpendicular bisectors is the circumcenter.

The radius of the circle is the distance from the circumcenter to any one of the three vertices.

Now measure the distance from the point in question to the circumcenter. If the distance is less than the radius the point is inside the circle. If it is greater then it is outside.

Answer 3

You know all the vertices of the triangle so you can easily form the equation of the circle. From th equation of the circle calculate the radius and center of the circle.
Now calculate the distance between the centerand the point
If the distance is greater than the radius, the poin lies outside
if distance is less than the radius, the point lies inside
if distance = radius, the point lies on the circle

Answer 4

ok, i think of i will do it utilizing geometry, some algebra, and probable some easy trig. No calculus. One situation is that i'm not particularly particular merely what section you prefer to discover, yet i've got sketched it out, and we are going to take it from there. till now we initiate, there's a theorem to put in the back of our heads: "The degree of an perspective shaped by utilizing 2 tangents that intersect interior the outdoors of a circle is one-a million/2 the version of the measures of the intercepted arcs." i don't be attentive to if we are going to use that or not, yet we could communicate approximately it. we are going to use your occasion of circles with radii 2 and 3. initiate by utilizing drawing a protracted horizontal line (the x axis). someplace interior the midst of the website, draw a circle of radius 2, tangent to the x-axis. Label the middle P and the element of tangency A. To the desirable of that circle, draw yet another, tangent to the 1st and to the x-axis, of radius 3. Label the middle Q and the element of tangency B. Draw PQ connecting the centers, and make larger PQ to the place it intersects the x-axis. Label that intersection O; draw the y-axis with the aid of O, and observe that O has the coordinates (0,0). Draw PA and QB, the two perpendicular to the x-axis and parallel to a minimum of one yet another. we could discover the area AB, and we are able to accomplish that utilizing the Pythagorean Theorem. all of us be attentive to PQ=5, PA=2, and QB=3, so AB^2 = PQ^2 - (QB-PA)^2 = 25 - a million = 24. hence, AB = sqrt(24) = 2 sqrt(6). The quadrilateral APQB is a trapezoid. (PA and QB are parallel.) AB is the altitude, because of the fact that AB is perpendicular to the two PA and QB. the part of trapezoid APQB is (a million/2)(AB)(PA+QB) = (a million/2)(2 sqrt(6))(2+3) = 5 sqrt(6). i'm going to supply up there, because of the fact i don't be attentive to what you prefer to discover. To proceed, i think of you will discover the coordinates of P and Q, and the area OA. you additionally can get the measures of angles POA and OPA, and you observe that triangles OPA and OQB are comparable. In different words, there is a lot you're able to do now in case you prefer to. this could've helped.

Answer 5

let equation of circle be
x^2+y^2+2gx+2fy+c=0
find g,f,c as this circle passes through points of triangle
(put those points in it and solve three equations)

after this for that point find the value of above expresion
if it comes out negative then the point lies inside circle