Let Q be a point on the circle B: x^2+y^2=a^2 | vedclass

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

Question

(D) Let $R(\alpha, \beta)$ be the required point,$Q(x_0, y_0)$ be a point on the circle,and $P(h, k)$ be a fixed point.Using the section formula,the c

Vedclass · Accepted Answer

$\left(\frac{q h}{p+q}, \frac{q k}{p+q}\right)$

Vedclass · Answer

(D) Let $R(\alpha, \beta)$ be the required point,$Q(x_0, y_0)$ be a point on the circle,and $P(h, k)$ be a fixed point.Using the section formula,the coordinates of $R$ are:$(\alpha, \beta) = \left(\frac{p x_0 + q h}{p+q}, \frac{p y_0 + q k}{p+q}\right)$From this,we get:$x_0 = \frac{(p+q)\alpha - qh}{p}$ and $y_0 = \frac{(p+q)\beta - qk}{p}$Since $Q(x_0, y_0)$ lies on the circle $x^2 + y^2 = a^2$,we substitute $x_0$ and $y_0$:$\left(\frac{(p+q)\alpha - qh}{p}\right)^2 + \left(\frac{(p+q)\beta - qk}{p}\right)^2 = a^2$Dividing by $\frac{(p+q)^2}{p^2}$,we get:$\left(\alpha - \frac{qh}{p+q}\right)^2 + \left(\beta - \frac{qk}{p+q}\right)^2 = \frac{p^2 a^2}{(p+q)^2}$Thus,the locus of $R(x, y)$ is a circle with centre $\left(\frac{qh}{p+q}, \frac{qk}{p+q}\right)$.

Let $Q$ be a point on the circle $B: x^2+y^2=a^2$ and $P(h, k)$ be a fixed point. If the locus of the point which divides the join of $P$ and $Q$ in the ratio $p: q$ is a circle $C$,then the centre of $C$ is

Similar Questions

The locus of the midpoints of the chords of the circle $x^2-2x+y^2=0$ drawn from the point $(0,0)$ on it is

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:

At which point on the line $x = 3$ are the tangents drawn to the circle $x^2 + y^2 = 8$ perpendicular to each other?

Let $A(5,4)$ and $B(5,-4)$ be two points. If $P(x,y)$ is a point in the coordinate plane such that $\angle APB = \frac{\pi}{4}$,then the point $P$ lies on the curve

If $A (c, 0)$ and $B (-c, 0)$ are two points,find the locus of point $P$ such that $PA^{2} + PB^{2} = AB^{2}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module