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

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.Mathematics

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