Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Let the tangent to the circle $\mathrm{C}_{1}: x^{2}+y^{2}=2$ at the point $\mathrm{M}(-1,1)$ intersect the circle $\mathrm{C}_{2}:(x-3)^{2}+(y-2)^{2}=5$ , at two distinct points $\mathrm{A}$ and $\mathrm{B}$ . If the tangents to $\mathrm{C}_{2}$ at the points $\mathrm{A}$ and $\mathrm{B}$ intersect at N, then the area of the triangle ANB is equal to :
Option: 1
$\frac{1}{2}$

Option: 2
$\frac{2}{3}$

Option: 3
$\frac{1}{6}$

Option: 4
$\frac{5}{6}$

Answers (1)

best_answer

$\mathrm{Using \: T=0, equation \: of\: tangent \: at \: M(-1,1)\: to \: x^{2}+y^{2}=2 \: is}$

$\mathrm{x x_{1}+y y_{1}=2} \\$

$\mathrm{-x+y=2} \\$

$\mathrm{\Rightarrow y=x+2}$

$\mathrm{\text { CM }=\frac{|2-3-2|}{\sqrt{1+1}}=\frac{3}{\sqrt{2}}}\\$

$\mathrm{Now \cos \theta=\frac{C M}{B C}=\frac{3}{\sqrt{2} \cdot \sqrt{5}}=\frac{3}{\sqrt{10}}}$

$\mathrm{\text { And } \quad B M =\sqrt{C B^{2}-C M^{2}}} \\$

$\mathrm{=\sqrt{5-\frac{9}{2}} }\\$

$\mathrm{=\frac{1}{\sqrt{2}}}$

$\mathrm{Area \: of \: \triangle A N B=2 \: area\: of\: \triangle N M B }$

$\mathrm{=2 \times \frac{1}{2} \times M B \cdot M N }\\$

$\mathrm{=\left(\frac{1}{\sqrt{2}}\right)\left(\frac{M B}{\tan (90-\theta)}\right) }\\$

$\mathrm{=\frac{1}{2} \cdot \tan \theta} \\$

$\mathrm{=\frac{1}{2} \cdot \frac{1}{3}=\frac{1}{6}}$

Hence the correct answer is option 3.

Posted by

vinayak

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pK_b of ammonia solution is 4.75, the pH of the mixture will be :
Option: 1 3.75
Option: 2 4.75

Trending Articles/News

anam-khan

IIIT Allahabad cut-offs for BTech ECE, IT, IT-Business Informatics goes up in last 3 years; seat matrix

anam-khan

IIIT Hyderabad JEE Main cut-offs for BTech CSE, ECE of last 5 years; modes of admissions for dual degrees

anam-khan

JEE Mains 2025 paper 2 objection window closes tonight; final answer key soon

Latest Question