Browse by Stream
QnA
Home

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let the tangents at two points  and  on the circle&nbsp

Let the tangents at two points \mathrm{A} and \mathrm{B} on the circle \mathrm{x^{2}+y^{2}-4 x+3=0} meet at origin \mathrm{\mathrm{O}(0,0)}. Then the area of the triangle \mathrm{\text { OAB }} is:

Option: 1

\frac{3 \sqrt{3}}{2}


Option: 2

\frac{3 \sqrt{3}}{4}


Option: 3

\frac{3}{2 \sqrt{3}}


Option: 4

\frac{3}{4 \sqrt{3}}


Answers (1)

best_answer

\begin{aligned} &\mathrm{r=\sqrt{4-3}=1}\\ &\mathrm{O \hat{A}=\sqrt{s_1}=\sqrt{3}}\\ & \mathrm{\therefore \tan \theta=\frac{1}{\sqrt{3}}}\\ &\text { Area of O A B}=\mathrm{2 \times\left(\frac{1}{2} \times \sqrt{3} \cos \theta \cdot \sqrt{3} \sin \theta\right)}\\ &=\mathrm{3 \sin \theta \cdot \cos \theta}\\ &=\mathrm{3 \times \frac{1}{2} \times \frac{\sqrt{3}}{2}}\\ &=\mathrm{\frac{3 \sqrt{3}}{4}}\\ &\therefore \text { Option (B) } \end{aligned}

Posted by

manish

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 session 2 exam date revised due to clash with Class 12 board exams 2025; new dates here
anam-khan

JEE Main Admit Card 2025 (OUT) LIVE: Session 2 hall ticket at jeemain.nta.nic.in; cut-offs, exam pattern
anam-khan

JEE Main 2025: BTech ECE cut-offs for female candidates go up at top NITs

Latest Question