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प्रश्न
The equation of the circle having centre (1, –2) and passing through the point of intersection of the lines 3x + y = 14 and 2x + 5y = 18 is ______.
पर्याय
x2 + y2 – 2x + 4y – 20 = 0
x2 + y2 – 2x – 4y – 20 = 0
x2 + y2 + 2x – 4y – 20 = 0
x2 + y2 + 2x + 4y – 20 = 0
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उत्तर
The equation of the circle having centre (1, –2) and passing through the point of intersection of the lines 3x + y = 14 and 2x + 5y = 18 is x2 + y2 – 2x + 4y – 20 = 0.
Explanation:
The point of intersection of 3x + y – 14 = 0 and 2x + 5y – 18 = 0 are x = 4, y = 2
i.e., The point (4, 2)
Therefore, the radius is = `sqrt(9 + 16)` = 5
And hence the equation of the circle is given by (x – 1)2 + (y + 2)2 = 25
or x2 + y2 – 2x + 4y – 20 = 0.
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