If x^{2}+y^{2}-2x+6y+10=0,then the value of ( | vedclass

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(D) Given equation: $x^{2}+y^{2}-2x+6y+10=0$Rearranging the terms to complete the squares:$(x^{2}-2x+1) + (y^{2}+6y+9) = 0$This can be written as:$(x-

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$10$

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(D) Given equation: $x^{2}+y^{2}-2x+6y+10=0$Rearranging the terms to complete the squares:$(x^{2}-2x+1) + (y^{2}+6y+9) = 0$This can be written as:$(x-1)^{2} + (y+3)^{2} = 0$Since the square of any real number is non-negative,the sum of two non-negative numbers can be zero only if each individual term is zero.Therefore:$(x-1)^{2} = 0 \implies x = 1$$(y+3)^{2} = 0 \implies y = -3$Now,calculate the value of $(x^{2}+y^{2})$:$x^{2}+y^{2} = (1)^{2} + (-3)^{2}$$x^{2}+y^{2} = 1 + 9 = 10$

If $x^{2}+y^{2}-2x+6y+10=0$,then the value of $(x^{2}+y^{2})$ is

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