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UPSC NDA II 2014 Mathematics Paper

For the next two (02) items that follow:

Consider the circles

x^2+y^2+2ax+c=0

and

x^2+y^2+2by+c=0

[NDA II 2014]

Question : 50 of 120

Marks: +1, -0

Solution:

x^2+y^2+2ax+c=0

\Rightarrow x^2+2ax+a^2+y^2=a^2-c

\Rightarrow (x+a)^2+y^2=a^2-c

Center

\equiv(-a,0)

and

\text{radius}^2=a^2-c

Similarly,

x^2+(y+b)^2=b^2-c

center

\equiv(0,-b)

and

\text{radius}^2=b^2-c

Distance between centers

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

The circle will touch if

Distance between centers = sum of radii

\Rightarrow \sqrt{a^2+b^2}=\sqrt{a^2-c^2}+\sqrt{b^2-c}

\Rightarrow a^2+b^2=a^2-c+b^2-c+2\sqrt{a^2-c}\sqrt{b^2-c}

\Rightarrow c=\sqrt{a^2-c}\sqrt{b^2-c}

\Rightarrow c^{2}=a^{2} b^{2}-a^{2} c-b^{2} c+c^{2}

\Rightarrow a^2b^2=c(a^2+b^2)

\Rightarrow\frac{1}{c}=\frac{1}{a^2}+\frac{1}{b^2}

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