If ...
the square of (the x-coordinate of the point minus the x-coordinate of the center of the circle)
added to
the square of (the y-coordinate of the point minus the y-coordinate of the center of the circle)
is equal to
the square of the circle's radius,
then the point is on the circle.
No, a tangent line cannot be constructed through a point A that lies inside circle C. A tangent line to a circle must touch the circle at exactly one point, and if point A is inside the circle, it cannot be tangent to any point on the circle. Instead, a line can be drawn from point A to the circle, but it will intersect the circle at two points rather than being tangent.
2,0
There are infinitely many such points. One of them is: (2,236, 4.291)
There are infinitely many points. One of these is (10, 0).
To prove that segments ( ab ) and ( ac ) are congruent in the construction of an equilateral triangle, you can use the property of circles. When you draw a circle with center ( a ) and radius ( ab ), point ( b ) lies on this circle. Similarly, if you draw a circle with center ( a ) and radius ( ac ), point ( c ) lies on this circle as well. Since both circles are constructed with the same radius from point ( a ), it follows that ( ab = ac ), proving that segments ( ab ) and ( ac ) are congruent.
True
No, a tangent line cannot be constructed through a point A that lies inside circle C. A tangent line to a circle must touch the circle at exactly one point, and if point A is inside the circle, it cannot be tangent to any point on the circle. Instead, a line can be drawn from point A to the circle, but it will intersect the circle at two points rather than being tangent.
2,0
Every diameter of the circle.
the artic circle
I'm not going to write the program for you, but the way to determine whether a point lies within a circle is very easy: just compare the distance between the point and the centerpoint of the circle with its radius. If the distance is smaller, it's inside the circle, if it's greater, then the point is outside.You can calculate the distance between the point and the centerpoint using Pythagoras's method. If the point is at (PX, PY) and the centerpoint is at (CX, CY), the distance can be calculated as such:DX = (CX - PX); // X distanceDY = (CY - PY); // Y distancedistance = sqrt( (DX * DX) + (DY * DY) );
There are infinitely many points. One of these is (10, 0).
There are infinitely many such points. One of them is: (2,236, 4.291)
Circle of Lies - 2012 is rated/received certificates of: Australia:MA15+ (2013)
To prove that segments ( ab ) and ( ac ) are congruent in the construction of an equilateral triangle, you can use the property of circles. When you draw a circle with center ( a ) and radius ( ab ), point ( b ) lies on this circle. Similarly, if you draw a circle with center ( a ) and radius ( ac ), point ( c ) lies on this circle as well. Since both circles are constructed with the same radius from point ( a ), it follows that ( ab = ac ), proving that segments ( ab ) and ( ac ) are congruent.
There are infinitely many points. One of these is (-7, 1)
To draw a great circle on a sphere, start by defining the diameter as the largest circle that can be drawn on the sphere's surface. For small circles, choose a point on the sphere and draw a circle with that point as the center and the radius less than the sphere's radius. Remember that the center of a small circle lies outside the circle on a sphere's surface.