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} = \( ( \sqrt{53})^2\) [which is form (x-h)^{2} +(y-k)^{2} = r^{2}]

Where h = 4, K= -5 and r = \( \sqrt{53}\)

The centre of the given circle is (4, -5) and its radius is \( \sqrt{53}\).

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