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Find the equation of the circle whose centre is (2, 3) and which passes through (1, 4). - Business Mathematics and Statistics

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प्रश्न

Find the equation of the circle whose centre is (2, 3) and which passes through (1, 4).

योग

उत्तर

Centre (h, k) = (2, 3)

Radius = `sqrt((1 - 2)^2 + (4 - 3)^2)`

`= sqrt((-1)^2 + 1^2)`

`= sqrt2`

Equation of the circle with centre (h, k) and radius r is (x – h)2 + (y – k)2 = r2

⇒ (x – 2)2 + (y – 3)2 = `(sqrt2)^2`

⇒ x2 – 4x + 4 + y2 – 6y + 9 = 2

⇒ x2 + y2 – 4x – 6y + 11 = 0

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Analytical Geometry - Exercise 3.4 [पृष्ठ ६४]

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सामाचीर कलवी Business Mathematics and Statistics [English] Class 11 TN Board
अध्याय 3 Analytical Geometry
Exercise 3.4 | Q 4 | पृष्ठ ६४
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