Home
Class 12
MATHS
In the diagram as shown, a circle is dra...


In the diagram as shown, a circle is drawn with centre `C(1,1)` and radius 1 and a line L. The line L is tangent to the circle at Q. Further L meets the y-axis at R and the x-axis at P in such a way that the angle OPQ equals `theta` where `0 lt theta lt (pi)/(2)`.
Equation of the line PR is

A

`x cos theta + y sin theta = sin theta + cos theta +1`

B

`x sin theta +y cos theta = cos theta + sin theta - 1`

C

`x sin theta +y cos theta = cos theta + sin theta +1`

D

`x tan theta +y = 1 +cot ((theta)/(2))`

Text Solution

Verified by Experts

The correct Answer is:
C

`m_(CQ) = cot theta`
`m_(PR) =- tan theta`
Equation of PR is
`y -(1+cos theta) =- tan theta (x-(1+sin theta))`
or `y +x theta =(1+cos theta) +tan theta (1+sin theta)`
`= (cos theta (1+cos theta) +sintheta (1+sin theta))/(cos theta)`
or `y +x tan theta = (1+cos theta +sin theta)/(cos theta)`
or `x sin theta +ycos theta = (cos theta + sin theta +1)`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • CIRCLES

    CENGAGE|Exercise Single Correct Answer Type|58 Videos
  • CIRCLES

    CENGAGE|Exercise Question Bank|32 Videos
  • CIRCLES

    CENGAGE|Exercise Question Bank|32 Videos
  • CIRCLE

    CENGAGE|Exercise JEE Advanced (Single Correct Answer Type)|14 Videos
  • COMPLEX NUMBERS

    CENGAGE|Exercise Question Bank|30 Videos

Similar Questions

Explore conceptually related problems

In the diagram as shown, a circle is drawn with centre C(1, 1) and radius I and a line L. The line Lis tangential to the circle at Q. Further L meet the y-axis at R and the x-axis at Pis such a way that the angle OPQ equals theta where 0 < theta

Line 2x+3y+1=0 is a tangent to the circle at (1,-1). This circle is orthogonal to a circle which is drawn having diameter as a line segment with end points (0,-1) and (-2,3). Then the equation of the circle is

Knowledge Check

  • In the diagram as shown, a circle is drawn with centre C(1,1) and radius 1 and a line L. The line L is tangent to the circle at Q. Further L meets the y-axis at R and the x-axis at P in such a way that the angle OPQ equals theta where 0 lt theta lt (pi)/(2) . Area of triangle OPR when theta = pi//4 is

    A
    `(3-2sqrt(2))`
    B
    `(3+2sqrt(2))`
    C
    `(6+4sqrt(2))`
    D
    none of these
  • A tangent to the hyperbola meets x-axis at P and y-axis at Q. Lines PR and QR are drawn such that OPRQ is a rectangle (where O is the origin) then R lies on

    A
    `(4)/(x^(2))+(2)/(y^(2))=1`
    B
    `(2)/(x^(2))-(4)/(y^(2))=1`
    C
    `(2)/(x^(2))+(4)/(y^(2))=1`
    D
    `(4)/(x^(2))-(2)/(y^(2))=1`
  • a tangent to the hyperbole X^2/4-y^2/2=1 meets x-axis at P and y-axis at Q. Lines PR and QR are deawn such that OPRQ is a rectangle (where O is the origin). Then R lies on :

    A
    `4/x^2+2/y^2=1`
    B
    `2/x^2-4/y^2=1`
    C
    `2/x^2+4/y^2=1`
    D
    `4/x^2-2/y^2=1`
  • Similar Questions

    Explore conceptually related problems

    The line 2x+3y=6, 2x+3y-8 cut the X-axis at A,B respectively. A line L=0 drawn through the point (2,2) meets the X-axis at C in such a way that abscissa of A,B,C are in arithmetic Progression. then the equation of the line L is

    The area of triangle formed by the lines l_1 and l_2 and the x-axis is:

    A tangent to the hyperbola (x^(2))/(4)-(y^(2))/(2)=1 1 meets x axis at p and y axis at Q line PR and QR are drawn such that OPRQ is rectangle where O is the origin then R lies on

    Equation of tangent to the circle x^(2)+y^(2)=50 at the point where the line x+7=0 meets the circle

    The lines 2x + 3y = 6, 2x + 3y = 8 cut the x-axis at A and B A line 'l' is drawn through the point (2,2) meets the x-axis at C in such a way that the abscissa of A, B and C are in geometric progression. The equation of the line 'l' is