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प्रश्न
A circle of area 9π square units has two of its diameters along the lines x + y = 5 and x – y = 1. Find the equation of the circle
उत्तर
Area of the circle = 9π
(i.e) πr2 = 9π
⇒ r2 = 9
⇒ r = 3
(i.e) radius of the circle = r = 3
The two diameters are x + y = 5 and x – y = 1
The point of intersection of the diameter is the centre of the circle = C
To find C: Solving x + y = 5 .......(1)
x – y = 1 ........(2)
(1) + (2)
⇒ 2x = 6
⇒ x = 3
Substituting x = 3 in (1) we get
3 + y = 5
⇒ y = 5 – 3 = 2
∴ Centre = (3, 2) and radius = 3
So equation of the circle is (x – 3)2 + (y – 2)2 = 32
(i.e) x2 + y2 – 6x – 4y + 4 = 0
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