Home
Class 12
MATHS
The equation of chord AB of the circle ...

The equation of chord AB of the circle `x^2+y^2=r^2` passing through the point P(1,1) such that `(PB)/(PA)=(sqrt2+r)/(sqrt2-r)`, `(0ltrltsqrt(2))`

Text Solution

Verified by Experts

The correct Answer is:
`y=x`
Chord through P(1,1) meets the circles at A and B.
`:. PA xx Pb=` ( length of tangent `)^(2) = PT ^(2)`
or `PA xx PB =2 -r^(2)` (1)
Given `(PB)/(PA)=(sqrt(2)+r)/(sqrt(2)-r)` (2)
Solving (1) and (2), we get
`PB =sqrt(2)+r,PA=sqrt(2)-r`
`implies AB=PB-PA=2r=` diameter of circle
Therefore, equation of chord AB is `y=x`
Promotional Banner

Topper's Solved these Questions

  • CIRCLE

    CENGAGE ENGLISH|Exercise CONCEPT APPLICATION EXERCISE 4.14|4 Videos

  • CIRCLE

    CENGAGE ENGLISH|Exercise CONCEPT APPLICATION EXERCISE 4.15|3 Videos

  • CIRCLE

    CENGAGE ENGLISH|Exercise CONCEPT APPLICATION EXERCISE 4.12|4 Videos

  • BINOMIAL THEOREM

    CENGAGE ENGLISH|Exercise Matrix|4 Videos

  • CIRCLES

    CENGAGE ENGLISH|Exercise Comprehension Type|8 Videos

Similar Questions

Explore conceptually related problems

Equation of chord AB of the circle x^2+y^2=2 passing through P(2,2) such that (PB)/(PA)=3, is given by (a) x=3y (b) x=y (c) y-2=sqrt3(x-2) (d) Non of these

Find the equation of the normal to the circle x^2+y^2=9 at the point (1/(sqrt(2)),1/(sqrt(2))) .

.The equation of the circle passing through three non-collinear points P( x_1,y_1 ) Q( x_2,y_2 ) and R( x_3,y_3 ) is

The polar of the point (2t,t-4) w.r.t. the circle x^(2)+y^(2)-4x-6y+1=0 passes through the point

The equation to the locus of the midpoints of chords of the circle x^(2)+y^(2)=r^(2) having a constant length 2l is

If the radius of the circle passing through the origin and touching the line x+y=2 at (1, 1) is r units, then the value of 3sqrt2r is

Find the equation of the plane through the points P(1,\ 1,\ 0),\ Q(1,\ 2,\ 1)a n d\ R(-2,\ 2,\ -1)dot

A variable chord of circle x^(2)+y^(2)+2gx+2fy+c=0 passes through the point P(x_(1),y_(1)) . Find the locus of the midpoint of the chord.

PQ is a chord of the circle x^(2)+y^(2)-2x-8=0 whose mid-point is (2, 2). The circle passing through P, Q and (1, 2) is

The mid point of the chord x-2y+7=0 w.r.t the circle x^(2)+y^(2)-2x-10y+1=0 is