Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • If a line, y=mx+c is a tangent to the circle,  and it is perpendicular to a line <img alt="L_{1}," src="https://l

If a line, y=mx+c is a tangent to the circle, (x-3)^{2}+y^{2}=1 and it is perpendicular to a line L_{1}, where, L_{1} is the tangent to the circle, x^{2}+y^{2}=1 at the point \left ( \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right ); ; then : 
Option: 1 c^{2}+7c+6=0
Option: 9 c^{2}-6c+7=0
Option: 17 c^{2}-7c+6=0
Option: 25 c^{2}+6c+7=0
 

Answers (1)

best_answer

 

 

Slope and Equation of Tangent -

Tangent and Normal

Slope and Equation of Tangent:

Let P(x0, y0) be a point on the continuous curve y = f(x), then the slope of the tangent to the curve at point P is

\left ( \frac{dy}{dx} \right )_{\left (x_0,y_0 \right )} .

\Rightarrow \left(\frac{d y}{d x}\right)_{\left ( x_0,y_0 \right )}=\tan \theta=\text { slope of tangent at } P

Where ? is the angle which the tangent at P (x makes with the positive direction of the x-axis as shown in the figure.

  • If the tangent is parallel to x-axis then ? = 00.

\\\Rightarrow\;\;\;\;\; \tan\theta=0\\\\\therefore\;\; \left (\frac{dy}{dx} \right )_{(x_0,y_0)}=0

  • If the tangent is perpendicular to x-axis then ? = 900

\\\Rightarrow\;\;\;\;\; \tan\theta\rightarrow \infty\;\;\;\;\text{or}\;\;\;\;\cot\theta=0\\\\\therefore\;\; \left (\frac{dx}{dy} \right )_{(x_0,y_0)}=0

Equation of Tangent:

Let the equation of curve y = f (x) and a point P (x0, y0)  lies on this curve.

The slope of the tangent to the curve at a point P is

\left ( \frac{dy}{dx} \right )_{\left (x_0,y_0 \right )} .

Hence, the equation of the tangent at point P is

(y-y_0)=\left ( \frac{dy}{dx} \right )_{(x_0,y_0)}\cdot(x-x_0)

Tangent from External Point:

If a point Q(a, b) does not lie on the curve y = f(x), then the equation of possible tangent to the curve y = f(x) (tangent passing through point Q (a, b)) can be found by solving point of contact P(x0, y0)  on the curve.  

 

\\P\;(x_0,y_0)\;\text{lies on the curve y = f(x), then}\\\\\text{\;\;\;\;\;\;\;}y_0=f(x_0)\\\\\text{Also, slope of PQ is}\\\\\text{\;\;\;\;\;\;}\frac{y_0-b}{x_0-a}=\left ( \frac{dy}{dx} \right )_{(x_0,y_0)}

By solving the above two equations we get point of contact point P and equation of tangent PQ.

-

 

 

Distance of a Point From a Line -

Distance of a point from a line

Perpendicular length from a point (x1,y1) to the line L : Ax + By + C = 0 is 

-

The slope of the tangent to x^{2}+y^{2}=1 \text { at }\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)

\begin{array}{l}{x^{2}+y^{2}=1} \\ {2 x+2 y y^{\prime}=0} \\ {y^{\prime}=-\frac{x}{y}=-1} \\ {y=m x+c \text { is tangent of } x^{2}+y^{2}=1}\end{array}

so m = 1

y = x + c

now distance of (3, 0) from y = x + c is

\begin{array}{l}{\left|\frac{c+3}{\sqrt{2}}\right|=1} \\ {c^{2}+6 c+9=2} \\ {c^{2}+6 c+7=0}\end{array}

Correct Option (4)

Posted by

vishal kumar

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Trending Articles/News

anam-khan

Engineering aspirant dies by suicide in Kota, fourth case this year
anam-khan

What is non-Aadhaar declaration in JEE Main 2025? Explain mismatch between image, face
anam-khan

Forgot JEE Main password 2025 for downloading admit card? Here’s what to do

Latest Question