If the point (lambda, 1+lambda) lies inside t | vedclass

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

Question

(C) For a point $(x_1, y_1)$ to lie inside the circle $x^2+y^2-r^2=0$,the condition is $x_1^2+y_1^2-r^2 < 0$. Given the point $(\lambda, 1+\lambda)$ a

Vedclass · Accepted Answer

$-1 < \lambda < 0$

Vedclass · Answer

(C) For a point $(x_1, y_1)$ to lie inside the circle $x^2+y^2-r^2=0$,the condition is $x_1^2+y_1^2-r^2 Given the point $(\lambda, 1+\lambda)$ and the circle $x^2+y^2-1=0$,we substitute the point into the equation: $\lambda^2 + (1+\lambda)^2 - 1 $\lambda^2 + 1 + 2\lambda + \lambda^2 - 1 $2\lambda^2 + 2\lambda $2\lambda(\lambda+1) Dividing by $2$,we get $\lambda(\lambda+1) This inequality holds when $\lambda$ lies between the roots $-1$ and $0$. Therefore,$-1 < \lambda < 0$.

If the point $(\lambda, 1+\lambda)$ lies inside the circle $x^2+y^2=1$,then

Similar Questions

Find the radius of a circle that touches the $y-$axis at point $P(0,2)$ and touches the circle $x^2 + y^2 = 16$ internally.

The equation of the incircle of the triangle formed by the lines $x=0, y=0$ and $3x+4y-24=0$ is

For the two circles $x^2 + y^2 = 16$ and $x^2 + y^2 - 2y = 0$,there is/are:

The point of contact of the circles $x^2+y^2+2x+2y+1=0$ and $x^2+y^2-2x+2y+1=0$ is

The number of common tangents that can be drawn to the circles $x^2+y^2=1$ and $x^2+y^2-2x-6y+6=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module