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