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# Find the Coordinates of Circumcentre and Radius of Circumcircle of ∆Abc If A(7, 1), B(3, 5) and C(2, 0) Are Given. - Geometry

ConceptConcepts of Coordinate Geometry

#### Question

Find the coordinates of circumcentre and radius of circumcircle of ∆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

$\Rightarrow {PA}^2 = {PB}^2 = {PC}^2$

$P A^2 = P B^2$

$\Rightarrow \left( 3 - a \right)^2 + \left( 5 - b \right)^2 = \left( 7 - a \right)^2 + \left( 1 - b \right)^2$

$\Rightarrow 9 + a^2 - 6a + 25 + b^2 - 10b = 49 + a^2 - 14a + 1 + b^2 - 2b$

$\Rightarrow a - b = 2 . . . . . \left( 1 \right)$

$P A^2 = P C^2$

$\Rightarrow \left( 7 - a \right)^2 + \left( 1 - b \right)^2 = \left( 2 - a \right)^2 + \left( 0 - b \right)^2$

$\Rightarrow 49 + a^2 - 14a + 1 + b^2 - 2b = 4 + a^2 - 4a + b^2$

$\Rightarrow 5a + b = 23 . . . . . \left( 2 \right)$

$\left( 1 \right) + \left( 2 \right)$

$a = \frac{25}{6}, b = \frac{13}{6}$

$= \sqrt{\left( \frac{25}{6} - 2 \right)^2 + \left( \frac{13}{6} - 0 \right)^2}$
$= \sqrt{\left( \frac{13}{6} \right)^2 + \left( \frac{13}{6} \right)^2}$
$= \frac{13}{6}\sqrt{2}$