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Conic Section

Short Answer Type

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

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Long Answer Type

12. Find the equation of a circle which passes through the points (2, 3) and (–1, 1) and whose centre lies on the straight line x – 3y – 11 = 0.

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13. Find the equation of the circle passing through the points A (5, 7), B (6, 6) and C (2, –2).

Let the equation of the circle be $space space space space left parenthesis straight x minus straight h right parenthesis squared plus left parenthesis straight y minus straight k right parenthesis squared space equals space straight r squared$ ...(i)
The circle passes through A (5, 7)
$rightwards double arrow$ $space space space left parenthesis 5 minus straight h right parenthesis squared plus left parenthesis 7 minus straight k right parenthesis squared space equals space straight r squared space rightwards double arrow space straight h squared plus straight k squared minus 10 straight h minus 14 straight k plus 25 plus 49 equals straight r squared$
$rightwards double arrow$ $space space space minus 10 straight h minus 14 straight k plus 74 space equals space straight r squared minus straight h squared minus straight k squared$ ...(ii)

The circle passes through B (6, 6)
$rightwards double arrow$ $left parenthesis 6 minus straight h right parenthesis squared plus left parenthesis 6 minus straight k right parenthesis squared space equals space straight r squared space rightwards double arrow space straight h squared plus straight k squared minus 12 straight h minus 12 straight k plus 36 plus 36 equals straight r squared$
$rightwards double arrow$ $negative 12 straight h minus 12 straight k plus 72 space equals space straight r squared minus straight h squared minus straight k squared$ ...(iii)
The circle passes through point C (2, -2)
$rightwards double arrow$ $space space space left parenthesis 2 minus straight h right parenthesis squared plus left parenthesis negative 2 minus straight k right parenthesis squared equals straight r squared space rightwards double arrow space straight h squared plus straight k squared minus 4 straight h plus 4 straight k plus 4 plus 4 equals straight r squared$
$rightwards double arrow$ $negative 4 straight h plus 4 straight k plus 8 equals straight r squared minus straight h squared minus straight k squared$ ...(iv)
Equating (ii) and (iii), we have -10h - 14k + 74 = -12h - 12k + 72
$rightwards double arrow$ 2h - 2k + 2 = 0 $rightwards double arrow$ h - k + 1 = 0 ...(v)
Equating (iii) and (iv), we have
-12h - 12k + 72 = -4h + 4k + 8 $rightwards double arrow$ 8h + 16k - 64 = 0 $rightwards double arrow$ h + 2k - 8 =0 ...(vi)
Subtracting (v) from (vi), we get
3k - 9 = 0 $rightwards double arrow$ k = 3
Also, from (vi), h + 2 (3) - 8 = 0 $rightwards double arrow$ h = 2
So, the centre of the circle is : $left parenthesis straight h comma space straight k right parenthesis space left right arrow space left parenthesis 2 comma space 3 right parenthesis$
It passes through point A (5, 7)
$rightwards double arrow$ r = distance between the centre and the point.
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Hence, the equation of the circle is: $space space left parenthesis straight x minus straight h right parenthesis squared plus left parenthesis straight y minus straight k right parenthesis squared space equals space straight r squared$
or $left parenthesis straight x minus 2 right parenthesis squared plus left parenthesis straight y minus 3 right parenthesis squared space equals space 5 squared space or space straight x squared plus straight y squared minus 4 straight x minus 6 straight y plus 4 plus 9 equals 25$
or $straight x squared plus straight y squared minus 4 straight x minus 6 straight y minus 12 equals 0$