Question

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Answer 1

Answer:

a^2 + b^2 -4a - 6b -3 = 0

Explanation:

It's so simple.....

a - 2 = √3 + √5

b - 3 = √3 - √5

(a - 2)^2 + (b - 3)^2 = (√3 + √5)^2 + (√3 - √5)^2

(a^2 - 4a + 4) + b^2 - 6b + 9 = 2[ (√3)^2 + (√5)^2]

a^2 + b^2 -4a - 6b + 13 = 16

a^2 + b^2 -4a - 6b -3 = 0

hence proved

Note :

(a+b)^2 + (a-b)^2 = 2[a^2+b^2] is used above for RHS