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Find the centre and radius of the circle x2 + y2 - 8x + 10y - 12 = 0
Solution:
The equation of the given circle is x2 + y2- 8x + 10y - 12 = 0
⇒ x2 + y2 - 8x + 10y -12 = 0
⇒ (x2 - 8x) + (y2 + 10y) = 12
⇒ {x2 - 2(x)(4) + 42} + {y2 + 2 (y)(5) + 52} - 16 - 25 = 12
⇒ (x - 4)2 + (y + 5)2 = 53
⇒ (x - 4)2 + |y - (- 5)|2 = (√53)2
which is of the form (x - h)2 + (y - k)2 = r2
Therefore, on comparing both equations we get
h = 4, k = - 5 and r = √53
Thus, the center of the given circle is (4, - 5) while its radius is √53
NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1 Question 8
Find the centre and radius of the circle x2 + y2 - 8x + 10y - 12 = 0
Summary:
The center and radius of the circle are (4, - 5) and √53 respectively
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