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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).

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Short Answer Type

14. Find the equation of the circle with radius 5, whose centre lies on x-axis and passing through the point (2, 3).

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15. Find the centre and radii of the following circle:

open parentheses straight x plus 3 over 2 close parentheses squared plus open parentheses straight y minus 5 over 2 close parentheses squared equals 289 over 4

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16. Find the centre and radii of the following circle:

left parenthesis straight x plus 5 right parenthesis squared plus left parenthesis straight y minus 3 right parenthesis squared space equals space 36

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17. Find the centre and radii of the following circle:

straight x squared plus straight y squared equals 49

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

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 $left parenthesis straight h comma space straight k right parenthesis space left right arrow space left parenthesis 4 comma space minus 5 right parenthesis comma space radius comma space straight r space equals space square root of 53$