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Question

The locus of point of intersection of the lines $${y}+ m{x} =\sqrt{\mathrm{a}^{2}{m}^{2}+{b}^{2}}$$ and $$m{y}-x=\sqrt{a^{2}+b^{2}{m}^{2}}$$, is


A
x2+y2=1a2+1b2
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B
x2+y2=a2+b2
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C
x2y2=a2b2
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D
1x2+1y2=a2b2
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Solution

The correct option is C $$x^{2}+y^{2}=a^{2}+b^{2}$$
Lets Assume both intersects at $$(h,k)$$, then this satisfies both lines.
$$k+m\ h=\sqrt{a^{2}m^{2}+b^{2}}$$
$$\Rightarrow km-h=\sqrt{a^{2}+b^{2}m^{2}}$$
$$\Rightarrow k^{2}+m^{2}h^{2}+2mhk=a^{2}\ m^{2}+b^{2}$$
$$\Rightarrow k^{2}m^{2}+h^{2}-2mhk=a^{2}+b^{2}m^{2}$$
$$\Rightarrow k^{2}(m^{2}+1)+h^{2}(1+m^{2})=a^{2}(1+m^{2})+b^{2}(1+m^{2})$$
$$\Rightarrow k^{2}+h^{2}=a^{2}+b^{2}$$
$$\Rightarrow x^{2}+y^{2}=a^{2}+b^{2}$$
Hence, option B is correct.

Maths

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