Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

 Tangents to the circle \mathrm{x^{2}+y^{2}=a^{2}} cut the circle \mathrm{x^{2}+y^{2}=2 a^{2}} at \mathrm{P} and \mathrm{Q}. The tangents at \mathrm{P} and \mathrm{Q} to the circle \mathrm{x^{2}+y^{2}=2 a^{2}} intersect at

Option: 1

right angles


Option: 2

60^{\circ}


Option: 3

 can't be determined


Option: 4

none of the above.


Answers (1)

best_answer

Equation of tangent at any point \mathrm{(a \cos \theta, a \sin \theta)\, is \, x \cos \theta+y \sin \theta=a}\quad \cdots(1)

Let the point of intersection of the tangents at P and Q be \mathrm{(h, k)}. If the tangents at P and Q intersect at right angles, then locus of \mathrm{(h, k)} will be director circle of \mathrm{x^{2}+y^{2}=2 a^{2}} i.e \mathrm{x^{2}+y^{2}=4 a^{2}}. PQ is chord of contact of the circle \mathrm{x^{2}+y^{2}=2 a^{2}} w.r.t. the point \mathrm{(h, k)} i.e. equation of \mathrm{P Q} is \mathrm{h x+k y=2 a^{2}}\quad \cdots(2)

(1) and (2) are same equation \frac{\cos \theta}{h}=\frac{\sin \theta}{k}=\frac{a}{2 a^{2}}=\frac{1}{2 a}$ $\cos ^{2} \theta+\sin ^{2} \theta=1 \Rightarrow h^{2}+k^{2}=4 a^{2}

\therefore locus of \mathrm{(h, k)}is \mathrm{x^{2}+y^{2}=4 a^{2}} which is director circle to \mathrm{x^{2}+y^{2}=2 a^{2}}

Hence (A) is the correct answer.

Posted by

Sanket Gandhi

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 session 2 in less than a month; how to crack ‘difficult, lengthy’ mathematics
anam-khan

Top engineering deemed universities in India for BTech admission with NIRF ranking
anam-khan

JEE Main 2025 session 2 application correction deadline today; edit fields at jeemain.nta.nic.in

Latest Question