Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A is (0, 0) and B is (3a, 0) and two points P and Q are taken on AB such that AP = PQ = QB. On AP, PQ and QB as diameters, three circles are drawn. The locus of a point S from which the tangents dr

A is (0, 0) and B is (3a, 0) and two points P and Q are taken on AB such that AP = PQ = QB. On AP, PQ and QB as diameters, three circles are drawn. The locus of a point S from which the tangents drawn to the three circles are such that sum of squares of the lengths of tangents is equal to b^{2} is
 

Option: 1

\mathrm{2 x^2+2 y^2-5 a x+6 a^2-b^2=0} 


Option: 2

\mathrm{x^2+y^2-3 a x+2 a^2-b^2=0}


Option: 3

\mathrm{3 x^2+3 y^2-9 a x+8 a^2-b^2=0}


Option: 4

none of these


Answers (1)

best_answer

From the given data, we obtain three circles, each with radius \frac{a}{2} and centres \mathrm{\left(\frac{a}{2}, 0\right)}

 

\mathrm{\left(\frac{3 a}{2}, 0\right) \text {, and }\left(\frac{5 a}{2}, 0\right) \text {. }} 

The equations of the three circles are

\mathrm{\begin{aligned} & C_1:\left(x-\frac{a}{2}\right)^2+y^2=\frac{a^2}{4} \\ & C_2:\left(x-\frac{3 a}{2}\right)^2+y^2=\frac{a^2}{4} \\ & C_3:\left(x-\frac{5 a}{2}\right)^2+y^2=\frac{a^2}{4} \end{aligned}}

Let S be (h, k). Then, the sum of squares of the lengths of tangents from S to the circles is 

\mathrm{\left[\left(h-\frac{a}{2}\right)^2+k^2\right]+\left[\left(h-\frac{3 a}{2}\right)^2+k^2\right]+\left[\left(h-\frac{5 a}{2}\right)^2+k^2\right]+\frac{3 a^2}{4}=b^2}

Or \mathrm{3\left(h^2+k^2\right)-9 a h+8 a^2=b^2}

Hence, the locus of S(h, k) is

\mathrm{3\left(x^2+y^2\right)-9 a x+8 a^2-b^2=0}

 

 

 

Posted by

Sayak

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question