If x,y∈Rsatisfies (x+5)2+(y−12)2−(14)2, then the minimum value of expression is −p Find p
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Solution
Let z=(x+5)2+(y−12)2−(14)2 Clearly minimum value of z will occur when first and second term will be zero, ⇒x=−5,y=12, Since square of a real number can be negative. Hence minimum value of z will be −(14)2=−196