Binomial Theorem

Question
CBSEENMA11015073

Find the equation of a circle having radius 5 units and the centre as the point of intersection of the straight lines 2x – y – 5 = 0 and 3x + 2y = 4.

Solution

Radius, r of the circle = 5                                                               ...(i)
The centre is the point of intersection of straight lines
                    2x - y - 5=0  and 3x + 2y = 4
Solving these two equations simultaneously, we have
                     4x - 2y = 10
                     3x + 2y = 4
Adding, we get 7x = 14 or x = 2
rightwards double arrow                 4 - y - 5 = 0           or   y = -1
∴ The co-ordinates of the centre are (2, -1)
rightwards double arrow                              h = 2, k = -1 and r = 5
The equation of the circle in the standard form 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 space space space space space space or space space left parenthesis straight x minus 2 right parenthesis space plus space left parenthesis straight y plus 1 right parenthesis squared space equals space left parenthesis 5 right parenthesis squared
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#6 {main}</pre>
which is the required equation of the circle.

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