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Tamil Nadu Board of Secondary EducationHSC Science Class 12

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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Question

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

Sum
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Solution

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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Chapter 5: Two Dimensional Analytical Geometry-II - Exercise 5.1 [Page 182]

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Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 12 TN Board
Chapter 5 Two Dimensional Analytical Geometry-II
Exercise 5.1 | Q 2 | Page 182
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