Maharashtra State Board course SSC (English Medium) Class 10th Board Exam
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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

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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

\[\Rightarrow {PA}^2 = {PB}^2 = {PC}^2\]

\[P A^2 = P B^2 \]

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

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

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

\[P A^2 = P C^2 \]

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

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

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

\[\left( 1 \right) + \left( 2 \right)\]

\[a = \frac{25}{6}, b = \frac{13}{6}\]

Radius = PC =

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

\[ = \sqrt{\left( \frac{13}{6} \right)^2 + \left( \frac{13}{6} \right)^2}\]

\[ = \frac{13}{6}\sqrt{2}\]

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Solution Find the Coordinates of Circumcentre and Radius of Circumcircle of ∆Abc If A(7, 1), B(3, 5) and C(2, 0) Are Given. Concept: Concepts of Coordinate Geometry.
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