Let Q(x_1, y_1) be a variable point and R(1, | vedclass

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

Question

(D) Let the coordinates of point $P$ be $(h, k)$.Since $P$ is the mid-point of $QR$,we have:$h = \frac{x_1 + 1}{2}$ and $k = \frac{y_1 + 0}{2}$This im

Vedclass · Accepted Answer

$x^2 + y^2 - x = 0$

Vedclass · Answer

(D) Let the coordinates of point $P$ be $(h, k)$.Since $P$ is the mid-point of $QR$,we have:$h = \frac{x_1 + 1}{2}$ and $k = \frac{y_1 + 0}{2}$This implies $x_1 = 2h - 1$ and $y_1 = 2k$.Since point $Q(x_1, y_1)$ lies on the circle $x^2 + y^2 = 1$,we substitute the values of $x_1$ and $y_1$ into the circle equation:$(2h - 1)^2 + (2k)^2 = 1$$4h^2 - 4h + 1 + 4k^2 = 1$$4h^2 + 4k^2 - 4h = 0$Dividing by $4$,we get $h^2 + k^2 - h = 0$.Replacing $(h, k)$ with $(x, y)$,the locus of point $P$ is $x^2 + y^2 - x = 0$.

Let $Q(x_1, y_1)$ be a variable point and $R(1, 0)$ be a point on the circle $x^2 + y^2 = 1$. If $P$ is the mid-point of $QR$,then the locus of the point $P$ is:

Similar Questions

If $A$ and $B$ are the points of intersection of the circle $x^2+y^2-8x=0$ and the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$,and a point $P$ moves on the line $2x-3y+4=0$,then the centroid of $\triangle PAB$ lies on the line:

If the tangents drawn from a point $P$ to the circle $x^2 + y^2 = a^2$ are perpendicular to the tangents drawn to the circle $x^2 + y^2 = b^2$,then the locus of $P$ is:

The locus of the moving point $P$,such that $2PA = 3PB$ where $A$ is $(0,0)$ and $B$ is $(4,-3)$,is

For any real number $t$,the point $\left(\frac{8 t}{1+t^2}, \frac{4\left(1-t^2\right)}{1+t^2}\right)$ lies on a/an

The locus of the intersection point of $x \cos \alpha - y \sin \alpha = a$ and $x \sin \alpha - y \cos \alpha = b$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module