Home
Class 12
MATHS
A line is drawn through a fix point P(al...

A line is drawn through a fix point P(`alpha, beta`) to cut the circle `x^2 + y^2 = r^2` at A and B. Then PA.PB is equal to :

Text Solution

Verified by Experts

From the figure,
PA.PB`=` Constant
Also, `PA.PB=PC^(2)`
But `PC^(2)=OP^(2)-OC^(2)`
`=alpha^(2)+beta^(2)-a^(2)`
Hence, `PA.PB=alpha^(2)+beta^(2)-a^(2)`
Promotional Banner

Topper's Solved these Questions

  • CIRCLE

    CENGAGE ENGLISH|Exercise Examples|13 Videos

  • CIRCLE

    CENGAGE ENGLISH|Exercise CONCEPT APPLICATION EXERCISE 4.1|1 Videos

  • BINOMIAL THEOREM

    CENGAGE ENGLISH|Exercise Matrix|4 Videos

  • CIRCLES

    CENGAGE ENGLISH|Exercise Comprehension Type|8 Videos

Similar Questions

Explore conceptually related problems

A line is drawn through the point P(3,11) to cut the circle x^(2)+y^(2)=9 at A and B. Then PA.PB is equal to

If a line is drawn through a point A(3,4) to cut the circle x^(2)+y^(2)=4 at P and Q then AP .AQ=

The tangent at the point (alpha, beta) to the circle x^2 + y^2 = r^2 cuts the axes of coordinates in A and B . Prove that the area of the triangle OAB is a/2 r^4/|alphabeta|, O being the origin.

If a line passes through the point P(1,-2) and cuts the x^2+y^2-x-y= 0 at A and B , then the maximum of PA+PB is

If (1+ax)^n = 1 + 8x + 24x^2 + … and a line through P(a, n) cuts the circle x^2 + y^2 = 4 in A and B , then PA.PB =

Let normals to the parabola y^(2)=4x at variable points P(t_(1)^(2), 2t_(1)) and Q(t_(2)^(2), 2t_(2)) meet at the point R(t^(2) 2t) , then the line joining P and Q always passes through a fixed point (alpha, beta) , then the value of |alpha+beta| is equal to

I If a point (alpha, beta) lies on the circle x^2 +y^2=1 then the locus of the point (3alpha.+2, beta), is

If a circle passes through the point (a, b) and cuts the circlex x^2+y^2=p^2 equation of the locus of its centre is

Comprehension (Q.6 to 8) A line is drawn through the point P(-1,2) meets the hyperbola x y=c^2 at the points A and B (Points A and B lie on the same side of P) and Q is a point on the lien segment AB. If the point Q is choosen such that PQ, PQ and PB are inAP, then locus of point Q is x+y(1+2x) (b) x=y(1+x) 2x=y(1+2x) (d) 2x=y(1+x)

Through a fixed point (h, k) secants are drawn to the circle x^2 +y^2 = r^2 . Then the locus of the mid-points of the secants by the circle is