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Question

The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line $$4x-5y=20$$ to the circle $$x^2+y^2=9$$, is 


A
20(x2+y2)36x+45y=0
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B
20(x2+y2)+36x45y=0
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C
36(x2+y2)20x+45y=0
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D
36(x2+y2)+20x45y=0
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Solution

The correct option is A $$20(x^2+y^2)-36x+45y=0$$

Consider a line  $$\alpha x+\beta y=9$$    ……. (1)     ( to be a circle)

But equation of circle is given by

$${{x}^{2}}+{{y}^{2}}=9\,\,\,\,\,\,\,......\left( 2 \right)$$

Let the midpoint $$(h, k)$$ at the circle  $$S$$.

Then by equation (1) and (2) to,

$$xh+yk={{h}^{2}}+{{k}^{2}}=9$$    …….(3)

Comparing equation  (1)  and (3)  to,  we get

$$ \alpha =\dfrac{9h}{{{h}^{2}}+{{k}^{2}}}\,\,\,\,\,\,\,\,\,\,and\,\,\,\,\,\,\beta =\dfrac{9h}{{{h}^{2}}+{{k}^{2}}} $$

$$ \,\,\,\,\, $$

Sine $$\alpha \,\,and\,\,\beta $$ lies on the given line of Line

$$4x-5y=20$$

Then,

$$4\alpha -5\beta =20$$     ……(4)

Put the value of  $$\alpha \,\,and\,\,\beta $$ in equation (4),

$$ \dfrac{4\times 9h}{{{h}^{2}}+{{k}^{2}}}-\dfrac{5\times 9h}{{{h}^{2}}+{{k}^{2}}}=20 $$

$$ 36h-45k=20\left( {{h}^{2}}+{{k}^{2}} \right) $$

$$ 20\left( {{h}^{2}}+{{k}^{2}} \right)-36h+45k=0 $$

The locus is a circle

$$20\left( {{x}^{2}}+{{y}^{2}} \right)-36x+45y=0$$

Option (A) is correct.


1000115_1059333_ans_ff6734fd57594c189855f62df9b3c8da.JPG

Mathematics

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