Hushar Mulga
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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 \[P\left( a, b \right)\].

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)