Find the coordinates of circumcentre and radius of circumcircle of D ABC if A(7, 1), B(3, 5) and C(2, 0) are given
16. Find the coordinates of circumcentre and radius of circumcircle of D ABC if A(7, 1), B(3, 5) and C(2, 0) are given
Solution
Let the circumcentre be [Pleft( a, b right)].
The given points are A(7, 1), B(3, 5) and C(2, 0).
The circumcircle passes through the points A, B and C and the thus,
PA = PB = PC
[Rightarrow {PA}^2 = {PB}^2 = {PC}^2]
[P A^2 = P B^2 ]
[ Rightarrow left( 3 – a right)^2 + left( 5 – b right)^2 = left( 7 – a right)^2 + left( 1 – b right)^2 ]
[ Rightarrow 9 + a^2 – 6a + 25 + b^2 – 10b = 49 + a^2 – 14a + 1 + b^2 – 2b]
[ Rightarrow a – b = 2 . . . . . left( 1 right)]
[P A^2 = P C^2 ]
[ Rightarrow left( 7 – a right)^2 + left( 1 – b right)^2 = left( 2 – a right)^2 + left( 0 – b right)^2 ]
[ Rightarrow 49 + a^2 – 14a + 1 + b^2 – 2b = 4 + a^2 – 4a + b^2 ]
[ Rightarrow 5a + b = 23 . . . . . left( 2 right)]
[left( 1 right) + left( 2 right)]
[a = frac{25}{6}, b = frac{13}{6}]
Radius = PC =
[= sqrt{left( frac{25}{6} – 2 right)^2 + left( frac{13}{6} – 0 right)^2}]
[ = sqrt{left( frac{13}{6} right)^2 + left( frac{13}{6} right)^2}]
[ = frac{13}{6}sqrt{2}]
Explanation:-
Given points are A(7,1), B(3,5), and C(2,0). Let the circumcenter be P(a,b).
Since the circumcircle passes through A, B, and C, the distance of PA, PB, and PC to point P will be the same.
Therefore, we have:
PA = PB = PC
Squaring both sides, we get:
PA^2 = PB^2 = PC^2
Using the distance formula, we get:
PA^2 = (7 – a)^2 + (1 – b)^2 PB^2 = (3 – a)^2 + (5 – b)^2 PC^2 = (2 – a)^2 + b^2
Since PA^2 = PB^2 and PB^2 = PC^2, we have:
(7 – a)^2 + (1 – b)^2 = (3 – a)^2 + (5 – b)^2 (3 – a)^2 + (5 – b)^2 = (2 – a)^2 + b^2
Expanding and simplifying, we get:
-6a + 24 = -10b + 26 -10a + 26 = -10b + 34
Solving for a and b, we get:
a = 25/6 b = 13/6
The circumradius of a triangle is the distance between the circumcenter and any of the vertices of the triangle. We can use the distance formula to find the distance between P and C:
PC = sqrt((2 – a)^2 + b^2)
Substituting the values of a and b, we get:
PC = sqrt((2 – 25/6)^2 + (13/6)^2) PC = sqrt((13/6)^2 + (13/6)^2) PC = 13/6 * sqrt(2)
Therefore, the circumradius of the triangle is 13/6 * sqrt(2).
Chapter 5. Co-ordinate Geometry – Problem set 5 (Page 122)