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Equation of the Tangent and Normal to the Circle - Practice Questions & MCQ

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

Quick Facts

• Equation of the Tangent in Point Form is considered one of the most asked concept.

• 68 Questions around this concept.

Solve by difficulty

Equation of the tangent to the circle, at the point (1, −1), whose centre is the point of intersection of the straight lines x − y = 1 and 2x + y = 3 is :

The range of values of $\lambda$ for which the circle $\mathrm{x^{2}+y^{2}=4}$ and $\mathrm{x^{2}+y^{2}-4 \lambda x+9=0}$ have two common tangents is

The number of common tangents that can be drawn to the circle $\mathrm{x^{2}+y^{2}-4 x-6 y-3=0\: and \: x^{2}+y^{2}+2 x+2 y+1=0}$ is

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Equation of the Tangent in Point Form

Equation of the Tangent in Point Form

Point Form

The equation of the tangent to a circle $\\\mathrm{x^2+y^2+2gx+2fy+c=0}$  at the point P(x1,y1) is  $\\\mathrm{xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0}$

Proof:

$\\\mathrm{C\;(-g,-f)\;is\;the\;centre \;of\;the\;circle}\\\text{As point P}(x_1,y_1)\text{ lies on the circle.}\\\\\mathrm{\therefore\;\;\;\;\;\;\;\;\;\;\;\; Slope\;of\;CP=\frac{y_1-(-f)}{x_1-(-g)}=\frac{y_1+f}{x_1+g}}\\\\\text{Here, PT is the perpendicular to CP.}\\\\\text{Thus,\;\;\;\;\;\;\;slope\;of\;PT = }\mathrm{-\left (\frac{x_1+g}{y_1+f} \right )}\\\\\text{Hence, the equation of the tangent at P}\;(x_1,y_1)\text{ is}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left(y-y_{1}\right)=-\left (\frac{x_{1}+g}{y_{1}+f} \right )\left(x-x_{1}\right)}\\\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\left ( y-y_1 \right )\left ( y_1+f \right )+\left ( x_1+g \right )\left ( x-x_1 \right )=0}\\\\\mathrm{\Rightarrow \;\;\;\;\;\;\;xx_1+yy_1+gx+fy=x_1^2+y_1^2+gx_1+fy_1}\\\\\text{now add }\mathrm{gx_1+fy_1+c}\text{ both side, we get}\\\\\mathrm{\Rightarrow \;\;\;\;\;\;\;xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=x_1^2+y_1^2+2gx_1+2fy_1+c}$

$\\\mathrm{i.e. \;\;\;\;\;\;\;xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0}\\(\mathrm{As,\;point\; P(x_1,y_1)\;lies \;on\;the\;circle\; so,\;\;x_1^2+y_1^2+2gx_1+2fy_1+c=0})$

NOTE:

In order to find out the equation of a tangent to any 2nd-degree curve, the following points must be kept in mind:

$\\ {x^{2} \text { is replaced by } x x_{1}} \\ {y^{2} \text { is replaced by } y y_{1}} \\ {x y \text { is replaced by } \frac{x y_{1}+x_{1} y}{2}} \\ {x \text { is replaced by } \frac{x+x_{1}}{2}} \\ {y \text { is replaced by } \frac{y+y_{1}}{2}}$

and c will remain c.

This method is applicable only for a 2nd degree conic.

Equation of Tangent in Parametric and Slope Forms

Equation of Tangent of Circle in Parametric Form

The equation of the tangent at the point $(\mathrm{a\cos\theta,a\sin\theta)}$ to a circle $\mathrm{x^2+y^2=a^2}$ is $\mathbf{x\cos\theta+y\sin\theta=a}$

Proof:

$\text { If } \mathrm{S}=\mathrm{x}^{2}+\mathrm{y}^{2}-\mathrm{a}^{2}=0 \text { is the circle, then the tangent at }\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { is } \mathrm{T}_{1}=\mathrm{x} \mathrm{x}_{1}+\mathrm{yy}_{1}-\mathrm{a}^{2}=0$

$\begin{array}{l}{\text { put }, \mathrm{x}_{1}=\mathrm{a} \cos \theta, \quad \mathrm{y}_{1}=\mathrm{a} \sin \theta} \\ {\text { we get }, \quad \mathrm{x} \cos \theta+\mathrm{y} \sin \theta=\mathrm{a}}\end{array}$

Slope Form

The equation of the tangent to a circle $\mathrm{x^2+y^2=a^2}$ having slope m is $\mathbf{y}=\mathbf{m} \mathbf{x} \pm \mathbf{a} \sqrt{\left(\mathbf{1}+\mathbf{m}^{2}\right)}$, and point of tangency is $\left(\pm \frac{a m}{\sqrt{\left(1+m^{2}\right)}}, \mp \frac{a}{\sqrt{\left(1+m^{2}\right)}}\right)$.

$\\\mathrm{Let\;\;y=mx+c\;\;be \;a\;tangent\;to\;the\;circle\;x^2+y^2=a^2.}\\ {\therefore \quad \text { Length of perpendicular from centre of circle }(0,0)} \\ {\text { on }(y=m x+c)=\text { radius of circle }}\\\mathrm{\therefore \;\;\;\;\;\;\;\;\;\;\;\frac{|c|}{\sqrt{1+m^{2}}}=a \Rightarrow c=\pm a \sqrt{1+m^{2}}}\\\\ {\text { substituting this value of } c \text { in } y=m x+c, \text { we get } \mathbf{y=m x\pm a\sqrt{(1+m^2)}}} \\ \text { which are the required equations of tangents. }$

$\begin{array}{l}{ \mathbf{Corollary}\text { : It also follows that } y=m x+c \text { is a tangent }} \\ {\text { to } x^{2}+y^{2}=a^{2} \text { if } c^{2}=a^{2}\left(1+m^{2}\right) \text { which is the condition of }} \\ {\text { tangency. }}\end{array}$

Point of Contact:

$\\ \text { Solving } x^{2}+y^{2}=a^{2} \text { and } y=m x \pm a \sqrt{1+m^{2}}, \text { simultaneously }\\\text{we get,}\\\\ {x=\pm \frac{a m}{\sqrt{\left(1+m^{2}\right)}}} \\ {y=\mp \frac{a}{\sqrt{\left(1+m^{2}\right)}}}\\\\ {\text { Thus, the coordinates of the points of contact are }} \\ {\left( \pm \frac{a m}{\sqrt{\left(1+m^{2}\right)}}, \mp \frac{a}{\sqrt{\left(1+m^{2}\right)}}\right)}$

NOTE:

$\\ \text{Equation of tangent of the circle\;} (\mathrm{x}-\mathrm{h})^{2}+(\mathrm{y}-\mathrm{k})^{2}=\mathrm{a}^{2} \text{ in terms of slope is }\\(\mathrm{y}-\mathrm{k})=\mathrm{m}(\mathrm{x}-\mathrm{h}) \pm \mathrm{a} \sqrt{\left(1+\mathrm{m}^{2}\right)}.$

Equation of the Normal to a Circle

Equation of the Normal to a Circle

A line passing through a point P on the curve which is perpendicular to the tangent at P is called the normal to the curve at P.

For a circle, the normal always passes through the centre of the circle.

Point Form:

The equation of the Normal at the point P(x1,y1) to a circle $\mathrm{S}=\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0$ is

$\mathbf{\frac{x-x_1}{g+x_1}=\frac{y-y_1}{f+y_1}}$

Proof:

As we know that the normal always passes through the centre C(-g, -f) of a circle.

Thus, the equation of the normal at point P to the circle

$\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;y-y_{1}=\frac{y_{1}+f}{x_{1}+g}\left(x-x_{1}\right)}$

$\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\frac{x-x_{1}}{x_{1}+g}=\frac{y-y_{1}}{y_{1}+f}}\\\\\text{This is the equation of the normal (CP) at point P of }\\\text{the circle}$

Tangent from a Point to the Circle

Tangent from a Point to the Circle

• If a point lies outside of a circle (here point is P), then two tangents can be drawn from P to the circle. Here, PQ and PR are two tangents.

• If a point lies on the circle, then one tangent can be drawn from the point to the circle. If C is the point, then ACB is the tangent

• If a point lies inside the circle, then no tangent can be drawn from the point to the circle.

To get equation of the tangents from an external point

$\begin{array}{l}{\text { Circle is }: \mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{a}^{2} \text { and let the tangent to it be : } \mathrm{y}=\mathrm{mx}+\mathrm{a} \sqrt{\left(1+\mathrm{m}^{2}\right)}} \\ {\text { As the tangent passes through point } \mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { lying out side the circle }} \\ {\text { then, } \mathrm{y}_{1}=\mathrm{mx}_{1}+\mathrm{a} \sqrt{\left(1+\mathrm{m}^{2}\right)}} \\ {\left(\mathrm{y}_{1}-\mathrm{mx_1}\right)^{2}=\mathrm{a^2}\left(1+\mathrm{m}^{2}\right)} \\ {\text { or, }\left(\mathrm{x}_{1}^{2}-\mathrm{a}^{2}\right) \mathrm{m}^{2}-2 \mathrm{mx}_{1} \mathrm{y}_{1}+\mathrm{y}_{1}^{2}-\mathrm{a}^{2}}=0 \\ {\text { Which is quadratic equation in m which gives two value of } \mathrm{m} .}\end{array}$

The tangents are real, imaginary or coincidence that is depends on the value of the discriminant.

If we have real values of m, then we can find the equations of 2 tangents using these slopes and the point P.

Length of tangent (PT)  from a point to a circle

$\begin{array}{l}{\text { The length of the tangent from a point P }\left(x_{1}, y_{1}\right) \text { to the circle }} \\ {x^{2}+y^{2}+2 g x+2 f y+c=0 \text { is } \sqrt{x_{1}^{2}+y_{1}^{2}+2 g x_{1}+2 f y_{1}+c}}\end{array}$

$\\\mathrm{In\;\;\Delta PTC,\;\;PT^2=PC^2-CT^2}\\\text{Here coordinates of C are (-g, -f)}\\\mathrm{Hence,\;\;PT^2=\left (\sqrt{\left ( x_1+g \right )^2+\left ( y_1+f \right )^2} \right )^2-\left (\sqrt{g^2+f^2-c} \right )^2}\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;PT=\sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c}\;\;\;\;\;\;\;\;\;\;\;[\because CT=radius]}$
This expression can also be written as $PT=\sqrt{S_1}$

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