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# Find the centre and radius of the circle x^{2} + y^{2} - 8x + 10y - 12 = 0

**Solution:**

The equation of the given circle is x^{2} + y^{2}- 8x + 10y - 12 = 0

⇒ x^{2} + y^{2} - 8x + 10y -12 = 0

⇒ (x^{2} - 8x) + (y^{2} + 10y) = 12

⇒ {x^{2} - 2(x)(4) + 4^{2}} + {y^{2} + 2 (y)(5) + 5^{2}} - 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} = r^{2}

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 x^{2} + y^{2} - 8x + 10y - 12 = 0

**Summary:**

The center and radius of the circle are (4, - 5) and √53 respectively

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