The two points A and B in a plane are such th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $A = (a, 0)$,$B = (-a, 0)$,and $P = (x, y)$.Given the condition $\frac{PA}{PB} = k$,we have $\frac{PA^2}{PB^2} = k^2$.Substituting the coordin

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Let $A = (a, 0)$,$B = (-a, 0)$,and $P = (x, y)$.Given the condition $\frac{PA}{PB} = k$,we have $\frac{PA^2}{PB^2} = k^2$.Substituting the coordinates,we get $\frac{(x - a)^2 + y^2}{(x + a)^2 + y^2} = k^2$.$(x - a)^2 + y^2 = k^2((x + a)^2 + y^2)$.$(x^2 - 2ax + a^2 + y^2) = k^2(x^2 + 2ax + a^2 + y^2)$.$(x^2 + y^2)(1 - k^2) - 2ax(1 + k^2) + a^2(1 - k^2) = 0$.If $k = 1$,the equation becomes $-4ax = 0$,which represents a straight line (the perpendicular bisector of $AB$),not a circle.Therefore,for the locus to be a circle,$k$ must not be equal to $1$.

The two points $A$ and $B$ in a plane are such that for all points $P$ lying on a circle satisfying $\frac{PA}{PB} = k$,then $k$ will not be equal to:

Similar Questions

Let the locus of the midpoints of the chords of the circle $x^2+(y-1)^2=1$ drawn from the origin intersect the line $x+y=1$ at $P$ and $Q$. Then,the length of $PQ$ is:

The locus of the centre of a circle passing through $(a, b)$ and cutting orthogonally to the circle $x^2 + y^2 = p^2$ is

The locus of the centre of the circle touching the $x$-axis and passing through the point $(-1, 1)$ is

The locus of the middle points of chords of the circle $x^2 + y^2 - 2x - 6y - 10 = 0$ which pass through the origin is:

The locus of a point,such that the difference of the squares of the lengths of the tangents drawn from it to two given circles is constant,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module