Question

Find the equation of a circle, the two ends of whose diameter AB are A (3,–1) and B (5, 3).

Solution

Mid-point of AB is $open parentheses fraction numerator 3 plus 5 over denominator 2 end fraction comma space fraction numerator negative 1 plus 3 over denominator 2 end fraction close parentheses$
∴ Centre C of the circle is $left parenthesis straight h comma space straight k right parenthesis space left right arrow space left parenthesis 4 comma space 1 right parenthesis$
Radius of the circle = CA
$rightwards double arrow$ $straight r equals square root of left parenthesis 4 minus 3 right parenthesis squared plus left parenthesis 1 plus 1 right parenthesis squared end root space equals space square root of 1 plus 4 end root space equals space square root of 5$
Hence, the equation of circle is :
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$rightwards double arrow$ $left parenthesis straight x minus 4 right parenthesis squared plus left parenthesis straight y minus 1 right parenthesis squared equals left parenthesis square root of 5 right parenthesis squared space space or space space space left parenthesis straight x minus 4 right parenthesis squared plus left parenthesis straight y minus 1 right parenthesis squared equals 5$

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Binomial Theorem

Find the equation of a circle, the two ends of whose diameter AB are A (3,–1) and B (5, 3).

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