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प्रश्न
Find the equation of the circle having (1, −2) as its centre and passing through the intersection of the lines 3x + y = 14 and 2x + 5y = 18.
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उत्तर
Solving 3x + y = 14 and 2x + 5y = 18 we get
x = 4 and y = 2
The radius is equal to the distance between (1, −2) and (4, 2)
\[r = \sqrt{\left( 4 - 1 \right)^2 + \left( 2 + 2 \right)^2}\]
\[ = \sqrt{9 + 16}\]
\[ = 5\]
Now, the equation of the circle is given by
\[ \Rightarrow \left( x - 1 \right)^2 + \left( y + 2 \right)^2 = 25\]
\[ \Rightarrow x^2 + y^2 - 2x + 4y - 20 = 0\]
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