Definitions [3]
A circumcircle is a circle that passes through all three vertices of a triangle. The three vertices lie on the boundary of the circle.
The circumcenter is the center point of the circumcircle. It is the unique point where all three perpendicular bisectors of the triangle's sides meet.
- The circumcenter is equidistant from all three vertices of the triangle.
The circumradius is the radius of the circumcircle. It is the distance from the circumcenter to any vertex of the triangle.
Formulae [1]
The distance between P(x1, y1) and Q(x2, y2) is
\[\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]
The distance of a point P(x, y) from the origin is
\[\sqrt{x^2+y^2}\]
Theorems and Laws [2]
If the point (x, y) is equidistant from the points (a + b, b – a) and (a – b, a + b), prove that bx = ay.
Let P(x, y), Q(a + b, b – a) and R (a – b, a + b) be the given points. Then
PQ = PR ...(Given)
⇒ `sqrt({x - (a + b)}^2 + {y - (b - a)}^2) = sqrt({x - (a - b)}^2 + {y - (a + b)}^2`
⇒ `{x - (a + b)}^2 + {y - (b - a)}^2 = {x - (a - b)}^2 + {y - (a + b)}^2`
⇒ x2 – 2x(a + b) + (a + b)2 + y2 – 2y(b – a) + (b – a)2 = x2 + (a – b)2 – 2x(a – b) + y2 – 2(a + b) + (a + b)2
⇒ –2x(a + b) – 2y(b – a) = –2x(a – b) – 2y(a + b)
⇒ ax + bx + by – ay = ax – bx + ay + by
⇒ 2bx = 2ay
⇒ bx = ay
If a ≠ b ≠ 0, prove that the points (a, a2), (b, b2) (0, 0) will not be collinear.
Let A(a, a2), B(b, b2) and C(0, 0) be the coordinates of the given points.
We know that the area of triangle having vertices (x1, y1), (x2, y2) and (x3, y3) is ∣∣`1/2`[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]∣∣ square units.
So,
Area of ∆ABC
`= |1/2|a(b^2 - 0) + b(0 - a^2) + 0(a^2 - b^2)||`
`=|1/2(ab^2 - a^2b)|`
`= 1/2|ab(b -a)|`
`!= 0 (∵ a!= b != 0)`
Since the area of the triangle formed by the points (a, a2), (b, b2) and (0, 0) is not zero, so the given points are not collinear.
