Two tangents are drawn from a point P to circle x^{2}+y^{2}-2 x-4 y+4=0 , such that angle between th

English

Gujarati

Hindi

10-1.Circle and System of Circles

hard

Two tangents are drawn from a point $P$ to the circle $x^{2}+y^{2}-2 x-4 y+4=0$, such that the angle between these tangents is $\tan ^{-1}\left(\frac{12}{5}\right)$, where $\tan ^{-1}\left(\frac{12}{5}\right) \in(0, \pi)$. If the centre of the circle is denoted by $C$ and these tangents touch the circle at points $A$ and $B$, then the ratio of the areas of $\Delta PAB$ and $\Delta CAB$ is :

$11:4$

$9:4$

$3:1$

$2:1$

(JEE MAIN-2021)

Solution

$\tan \theta=\frac{12}{5}$

$PA =\cot \frac{\theta}{2}$

$\therefore$ area of $\Delta PAB =\frac{1}{2}( PA )^{2} \sin \theta=\frac{1}{2} \cot ^{2} \frac{\theta}{2} \sin \theta$

$=\frac{1}{2}\left(\frac{1+\cos \theta}{1-\cos \theta}\right) \sin \theta$

$=\frac{1}{2}\left(\frac{1+\frac{5}{13}}{1-\frac{5}{13}}\right)\left(\frac{12}{13}\right)=\frac{1}{2} \frac{18}{18} \times \frac{2}{13}=\frac{27}{26}$

area of $\Delta CAB =\frac{1}{2} \sin \theta=\frac{1}{2}\left(\frac{12}{13}\right)=\frac{6}{13}$

$\therefore \frac{\text { area of } \Delta PAB }{\text { area of } \Delta CAB }=\frac{9}{4}$

Standard 11

Mathematics

The line $2 x – y +1=0$ is a tangent to the circle at the point $(2,5)$ and the centre of the circle lies on $x-2 y=4$. Then, the radius of the circle is

hard

(JEE MAIN-2021)

Two tangents are drawn from the point $\mathrm{P}(-1,1)$ to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}-6 \mathrm{y}+6=0$. If these tangents touch the circle at points $A$ and $B$, and if $D$ is a point on the circle such that length of the segments $A B$ and $A D$ are equal, then the area of the triangle $A B D$ is eqaul to:

medium

(JEE MAIN-2021)

Number of integral points interior to the circle $x^2 + y^2 = 10$ from which exactly one real tangent can be drawn to the curve $\sqrt {{{\left( {x + 5\sqrt 2 } \right)}^2} + {y^2}} \, – \sqrt {{{\left( {x – 5\sqrt 2 } \right)}^2} + {y^2}\,} \, = 10$ are (where integral point $(x, y)$ means $x, y \in I)$

normal

The two circles which passes through $(0,a)$ and $(0, – a)$ and touch the line $y = mx + c$ will intersect each other at right angle, if

hard

A circle with centre $(a, b)$ passes through the origin. The equation of the tangent to the circle at the origin is

easy

Solution

Similar Questions

The line $2 x – y +1=0$ is a tangent to the circle at the point $(2,5)$ and the centre of the circle lies on $x-2 y=4$. Then, the radius of the circle is

The two circles which passes through $(0,a)$ and $(0, – a)$ and touch the line $y = mx + c$ will intersect each other at right angle, if

A circle with centre $(a, b)$ passes through the origin. The equation of the tangent to the circle at the origin is

Start a Free Trial Now