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UPSC NDA I 2016 Mathematics Solved Paper

Question Numbers: 72-73

Direction: For the next two (2) items that follow:

Consider a circle passing through the origin and the points (a, b) and (-b, -a)

Question : 72 of 120

Marks: +1, -0

On which line does the centre of the circle lie?

Solution:

Concept:

The centre of a circle with equation

x^{2}+y^{2}+2gx+2fy+c=0

(-g,-f)

Explanation:

The circle passes through

(0,0)

, so

c=0

Through

(a,b)

a^{2}+b^{2}+2ga+2fb=0\quad

(i)

Through

(-a,-b)

a^{2}+b^{2}-2ga-2fb=0\quad

(ii)

Multiply (i) by

a

and (ii) by

b

and add:

a(a^{2}+b^{2}+2ga+2fb)+b(a^{2}+b^{2}-2ga-2fb)=0

(a^{2}+b^{2})(a+b)+2ga^{2}-2gb^{2}=0

g=\frac{(a^{2}+b^{2})(a+b)}{2(b^{2}-a^{2})}=\frac{a^{2}+b^{2}}{2(b-a)}

Multiply (i) by

b

and (ii) by

a

and add:

b(a^{2}+b^{2}+2ga+2fb)+a(a^{2}+b^{2}-2ga-2fb)=0

(a^{2}+b^{2})(a+b)+2fb^{2}-2fa^{2}=0

f=\frac{(a^{2}+b^{2})(a+b)}{2(a^{2}-b^{2})}=\frac{a^{2}+b^{2}}{2(a-b)}

Centre

=(-g,-f)=\left(\frac{-(a^{2}+b^{2})}{2(b-a)},\frac{-(a^{2}+b^{2})}{2(a-b)}\right)=\left(\frac{a^{2}+b^{2}}{2(a-b)},\frac{a^{2}+b^{2}}{2(b-a)}\right)

Now check

x+y=0

\frac{a^{2}+b^{2}}{2(a-b)}+\frac{a^{2}+b^{2}}{2(b-a)}=\frac{a^{2}+b^{2}}{2(a-b)}-\frac{a^{2}+b^{2}}{2(a-b)}=0

Thus centre lies on

x+y=0

Answer:

Option A:

x+y=0

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