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Find the equation of the circle with centre (2, −1) and passing through the point (3, 6) in standard form - Mathematics

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प्रश्न

Find the equation of the circle with centre (2, −1) and passing through the point (3, 6) in standard form

बेरीज

उत्तर

Centre = C = (2, −1)

Passing through = A = (3, 6)

So radius = CA

= `sqrt((2 - 3)^2 + (- 1 - 6)^2)`

= `sqrt(1 + 49)`

= `5sqrt(50)`

Now centre = (2, −1) and radius = `sqrt(50)`

So equation of the circle is

(i.e) (x – 2)2 + (y + 1)2 = `sqrt(50)^2`

⇒ (x – 2)2 + (y + 1)2 = 50

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पाठ 5: Two Dimensional Analytical Geometry-II - Exercise 5.1 [पृष्ठ १८२]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
पाठ 5 Two Dimensional Analytical Geometry-II
Exercise 5.1 | Q 2 | पृष्ठ १८२

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