Binomial Theorem

Question
CBSEENMA11015079

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

Solution




The centre of the circle lies on x-axis.
rightwards double arrow  Its co-ordinates can be supposed as (h, 0)
The circle passes through point A (2, 3) and has a radius 5.
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rightwards double arrow       straight h squared minus 4 straight h minus 12 equals 0 space space rightwards double arrow space left parenthesis straight h minus 6 right parenthesis space left parenthesis straight h plus 2 right parenthesis space equals space 0 space rightwards double arrow space straight h space equals space 6 space space or space straight h space equals space minus 2
When h = 6, the centre of circle is left parenthesis straight h comma space 0 right parenthesis space left right arrow space left parenthesis 6 comma space 0 right parenthesis comma space radius of circle = 5
∴  The equation of circle is: left parenthesis straight x minus 6 right parenthesis squared plus left parenthesis straight y minus 0 right parenthesis squared space equals space 5 squared
or   straight x squared plus straight y squared minus 12 straight x plus 36 minus 25 equals 0 space space or space space straight x squared plus straight y squared minus 12 straight x plus 11 equals 0
When h = -2, the centre of circle is (-2, 0) and the radius is 5
∴ the equation of 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
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or          straight x squared plus straight y squared plus 4 straight x plus 4 equals 25 space space or space straight x squared plus straight y squared plus 4 straight x minus 21 equals 0
Hence, the equation of circle is: straight x squared plus straight y squared minus 12 straight x plus 11 equals 0 space or space straight x squared plus straight y squared plus 4 straight x minus 21 equals 0

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