Question

Find the centre and radii of the following circle:
$straight x squared plus straight y squared minus 8 straight x plus 10 straight y minus 12 equals 0$

Solution

The given equation of the circle is: $straight x squared plus straight y squared minus 8 straight x plus 10 straight y minus 12 equals 0$
$rightwards double arrow$ $left parenthesis straight x squared minus 8 straight x right parenthesis plus left parenthesis straight y squared plus 10 straight y right parenthesis equals 12 space space rightwards double arrow space space left square bracket straight x squared minus 8 straight x plus left parenthesis 4 right parenthesis squared right square bracket plus left square bracket straight y squared plus 10 straight y plus left parenthesis 5 right parenthesis squared right square bracket space equals space 12 plus left parenthesis 4 right parenthesis squared plus left parenthesis 5 right parenthesis squared$
$rightwards double arrow$ $space space space left parenthesis straight x minus 4 right parenthesis squared plus left parenthesis straight y plus 5 right parenthesis squared space equals space 53 space rightwards double arrow space left parenthesis straight x minus 4 right parenthesis squared plus left parenthesis straight y plus 5 right parenthesis squared space equals space left parenthesis square root of 53 right parenthesis squared$
Comparing it with the standard equation of circle, $left parenthesis straight x minus straight h right parenthesis squared plus left parenthesis straight y minus straight k right parenthesis squared equals straight r squared comma$ we get
$straight h equals 4 comma space straight k equals negative 5 comma space straight r equals square root of 53$
Hence, the co-ordinates of the centre are $left parenthesis straight h comma space straight k right parenthesis space left right arrow space left parenthesis 4 comma space minus 5 right parenthesis comma space radius comma space straight r space equals space square root of 53$

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Binomial Theorem

Find the centre and radii of the following circle:
$straight x squared plus straight y squared minus 8 straight x plus 10 straight y minus 12 equals 0$

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