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Chord of Contact - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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  • 37 Questions around this concept.

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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 $A$ and $B$. If the tangents to $C_2$ at the points $A$ and $B$ intersect at $N$, then the area of the triangle ANB is equal to :

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

The equation of the chord of contact of the origin w.r.t. circle \mathrm{ x^2+y^2-2 x-4 y-4=0} is
 

Equation of chord \mathrm{A B}  of circle \mathrm{x^{2}+y^{2}=2} passing through \mathrm{P(2,2)}  such that \mathrm{P B / P A=3}, is given by:

If chord of contacts are drawn to the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ from different points on the line $4 x+25 y=100$, then each of these chords pass through which point

The chords of contact of the pair of tangents drawn from each point on the line $2 \mathrm{x}+\mathrm{y}=4$ to the circle $\mathrm{x}^2+\mathrm{y}^2=1$ pass through a fixed point -

The equation of chord of contact of point $(2,-1)$ with respect to parabola $y=x^2-2 x+2$ is

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The locus of midpoints of the chords of the circle $x^2+y^2=4$ that pass through the point $(1,1)$ is

Equation of chord AB of circle $\mathrm{x}^2+\mathrm{y}^2=2$ passing through $\mathrm{P}(2,2)$ such that $\frac{\mathrm{PB} }{ \mathrm{PA}}=3$ , is given by-

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The point of intersection of tangents to $\frac{x^2}{16}+\frac{y^2}{9}=1$ at points where the line $x+8 y-8=0$ intersects it, is

 

 

Concepts Covered - 1

Chord Bisected at a Given Point

Equation of Chord Bisected at a Given Point 

The equation of the chord $(A B)$ of the circle $S=0$ whose midpoint is $M\left(x_1, y_1\right)$ is

coordinate of point O is $(0,0)$ and Point M is $\left(x_1, y_1\right)$
slope of $\mathrm{OM}=\frac{0-\mathrm{y}_1}{0-\mathrm{x}_1}=\frac{\mathrm{y}_1}{\mathrm{x}_1}$
$\therefore \quad$ slope of $\mathrm{AB}=-\frac{\mathrm{x}_1}{\mathrm{y}_1}$
the, equation of $A B$ is $y-y_1=-\frac{x_1}{y_1}\left(x-x_1\right)$
or

$
\mathrm{yy}_1-\mathrm{y}_1^2=\mathrm{xx}_1+\mathrm{x}_1^2
$

or

$
\mathrm{xx}_1+\mathrm{yy}_1-\mathrm{a}^2=\mathrm{x}_1^2+\mathrm{y}_1^2-\mathrm{a}^2 \quad \text { or } \quad \mathbf{T}=\mathbf{S}_1
$

Note:
Same result $T=S_1$, can also be applied to get chord bisected at a given point for any circle of type $x^2+y^2+2 g x+2 f y+c=0$ as well.

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Chord Bisected at a Given Point

Mathematics for Joint Entrance Examination JEE (Advanced) : Coordinate Geometry

Page No. : 4.23

Line : 8

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