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
Let the circumcentre be \[P\left( a, b \right)\].
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)