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Find the equation of the circle which passes through the points (20, 3), (19, 8) and (2, –9). Find its centre and radius. - Mathematics

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Question

Find the equation of the circle which passes through the points (20, 3), (19, 8) and (2, –9). Find its centre and radius.

Sum

Solution

By substitution of coordinates in the general equation of the circle given by x2 + y2 + 2gx + 2fy + c = 0

We have 40g + 6f + c = – 409

38g + 16f + c = – 425

4g – 18f + c = – 85

From these three equations, we get

g = – 7

f = – 3

And c = –111

Hence, the equation of the circle is

x2 + y2 – 14x – 6y – 111 = 0

or (x – 7)2 + (y – 3)2 = 132

Therefore, the centre of the circle is (7, 3) and radius is 13.

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Chapter 11: Conic Sections - Solved Examples [Page 195]

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NCERT Exemplar Mathematics [English] Class 11
Chapter 11 Conic Sections
Solved Examples | Q 7 | Page 195
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