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    JEE Main 2026 Session 2 Marks vs Percentile: Shift-wise Analysis, Expected Rank & Score Trends

    Pair of Tangent - Practice Questions & MCQ

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

    Quick Facts

    • 17 Questions around this concept.

    Solve by difficulty

    The tangents to \mathrm{x^{2}+y^{2}=a^{2}} having inclinations \mathrm{\alpha} and \mathrm{\beta } intersect at \mathrm{P }. If \mathrm{\cot \alpha+\cot \beta=0 }, then the locus of \mathrm{P } is

    If the two tangents are drawn on hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1} in such a way that the product of their gradients is \mathrm{c^2}, then they intersect on the curve:

    The angles between a pair of tangents drawn from a point P to the circle $\mathrm{x^2+y^2+4 x-6 y+9 \cos ^2 \alpha+13 \sin ^2 \alpha=0 \text { is } 2 \alpha \text {. }}$. The equation of the locus of the point P, is

    The angle between the two tangents from the origin to the circle $\mathrm{(x-7)^{2}+(y+1)^{2}=25}$ equals

    Concepts Covered - 1

    Pair of Tangent

    Pair of Tangent 

    The combined equation of the pair of tangents drawn from $P\left(x_1, y_1\right)$ to the circle $S: x^2+y^2=a^2$ is

    $
    \left(x^2+y^2-a^2\right)\left(x_1^2+y_1^2-a^2\right)=\left(x x_1+y y_1-a^2\right)^2 \quad \text { or } \quad S S_1=T^2
    $
    Where,

    $
    \begin{aligned}
    & S \equiv x^2+y^2-a^2 \\
    & S_1 \equiv x_1^2+y_1^2-a^2 \\
    & T \equiv x x_1+y y_1-a^2
    \end{aligned}
    $
    The combined equation of a pair of tangents drawn from $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ to a circle $x^2+y^2+2 g x+2 f y+c=0$ is

    $
    S S_1=T^2
    $

     

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    Pair of Tangent

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

    Page No. : 4.22

    Line : 40

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