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11. Find the equation of a circle, the two ends of whose diameter are A(4, –5) and R (–2, –1).

Let C be the centre of the circle
∴ C is the mid-point of AB.
∴ Co-ordinates of C, (h, k) $left right arrow$ $open parentheses fraction numerator 4 minus 2 over denominator 2 end fraction comma space fraction numerator negative 5 minus 1 over denominator 2 end fraction close parentheses$
$rightwards double arrow$ (h, k) $left right arrow$ (1, -3)
∴ h = 1, k = -3

Also, radius, r = CB
$rightwards double arrow$ $straight r equals square root of left parenthesis 1 plus 2 right parenthesis squared plus left parenthesis negative 3 plus 1 right parenthesis squared end root space equals space square root of 9 plus 4 end root space equals space square root of 13$
∴ The equation of circle is: $left parenthesis straight x minus straight h right parenthesis squared plus left parenthesis straight y minus straight k right parenthesis squared equals straight r squared space or space left parenthesis straight x minus 1 right parenthesis squared plus left parenthesis straight y plus 3 right parenthesis squared space equals space left parenthesis square root of 13 right parenthesis squared$
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Hence, the equation of the circle is : $space space straight x squared plus straight y squared minus 2 straight x plus 6 straight y minus 3 equals 0$