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Find the sum of coefficients of \mathrm{x \: and\: y} in the locus of the mid-point of the chord of the circle \mathrm{x^{2}+y^{2}=a^{2}} which subtend a right angle at the point \mathrm{(p,q)} .

Option: 1

\mathrm{p+q}


Option: 2

\mathrm{p-q}


Option: 3

\mathrm-{(p+q)}


Option: 4

None of these


Answers (1)

best_answer

Let the mid point of the chord be \mathrm{M}(\mathrm{h}, \mathrm{k}) and circle be represented by

\mathrm{S} \equiv \mathrm{x}^2+\mathrm{y}^2-\mathrm{a}^2=0

The equation of chord is

\mathrm{T}=\mathrm{S}_1
\mathrm{hx}+\mathrm{ky}-\mathrm{a}^2=\mathrm{h}^2+\mathrm{k}^2-\mathrm{a}^2          

or\mathrm{hx}+\mathrm{ky}=\mathrm{h}^2+\mathrm{k}^2  ............(i)

\mathrm{OM=}length of perpendicular from \mathrm{O(0,0)} to the line

(1)

\mathrm{ =\frac{h^2+k^2}{\sqrt{h^2+k^2}} }
\mathrm{ =\sqrt{\left(h^2+k^2\right)} }

In \mathrm{\quad \triangle O A M,(A M)^2=(O A)^2-(O M)^2}
\mathrm{ \therefore \quad(A M)^2=\left(a^2-h^2-k^2\right) }
Since \mathrm{ M } is the mid point of \mathrm{ A B }

\therefore \quad \mathrm{MA}=\mathrm{MB}=\mathrm{MP} (since distance from the mid point of hypotenuse of right angled triangle of the vertices are equal)

\therefore \quad(\mathrm{MA})^2=(\mathrm{MP})^2

\Rightarrow \quad \mathrm{a}^2-\mathrm{h}^2-\mathrm{k}^2=(\mathrm{h}-\mathrm{p})^2+(\mathrm{k}-\mathrm{q})^2

\Rightarrow \quad 2 \mathrm{~h}^2+2 \mathrm{k}^2-2 \mathrm{ph}-2 \mathrm{qk}+\mathrm{p}^2+\mathrm{q}^2-\mathrm{a}^2=0

Hence locus of mid point \mathrm{M}(\mathrm{h}, \mathrm{k}) is

\mathrm{ 2 x^2+2 y^2-2 p x-2 q y+p^2+q^2-a^2=0 }

or \mathrm{ \quad x^2+y^2-p x-q y+\frac{1}{2}\left(p^2+q^2-a^2\right)=0. }

Hence option 3 is correct.

Posted by

sudhir.kumar

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