Application of Derivatives

Given, equation of circle is
x2+y2=a2‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
Also P and Q are end points of diameter.
Let P≡(a‌cos‌θ,asin‌θ)
Q≡(−a‌cos‌θ,−asin‌θ)
According to the question,
s‌=‌

|a‌cos‌θ+asin‌θ−1|

√2

‌=‌

|a(cos‌θ+sin‌θ)−1|

√2

‌ and ‌t‌=‌

|−a(cos‌θ+sin‌θ)−1|

√2

=1
‌ Now, st ‌‌=‌

|1−a2(cos‌θ+sin‌θ)2|

2

‌=‌

|1−a2(1+sin‌2θ)|

2

So, ( st ) will be maximum, if ( 1+sin‌2θ ) minimum,
‌∵(sin‌2θ)min=−1(‌ i.e. ‌sin‌2θ≥−1)
‌⇒1+sin‌2θ≥0∴(1+sin‌2θ)min=0
‌∴(a2(1+sin‌2θ))min=0‌ and hence, ‌st=‌

1

2

‌‌ Also, ‌(a2(1+sin‌2θ))max=2a2
‌⇒‌‌st=‌

|1−2a2|

2

>‌

1

2

‌ for ‌a>‌

1

√2

So, (st)‌max ‌ occurs when
a2‌(1+sin‌2θ)=2a2
⇒1+sin‌2θ=2(∵a>‌

1

√2

∴|a|=a)
⇒cos‌θ+sin‌θ=√2
⇒θ=45∘
∴‌‌‌s=‌

|√2a−1|

√2

,‌ and ‌t=‌

∣√2a+1∣

√2

∴‌‌‌a>‌

1

√2

⇒√2a>1
∴‌‌‌s=‌

√2a−1

√2

,
‌t=‌

√2a+1

√2

=a+‌

1

√2

Clearly, t>s
∴‌‌(a+‌

1

√2

) is the final answer.