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

If the locus of the midpoints of the chords of the circle $x^2+y^2=25$,which subtend a right angle at the origin,is given by $\frac{x^2}{a^2}+\frac{y^2}{a^2}=1$,then $|a|=$

$A$ circle touches both the $y$-axis and the line $x+y=0$. Then the locus of its center is

The shortest distance between the curves $y^2 = x^3$ and $9x^2 + 9y^2 - 30y + 16 = 0$ is

The hypotenuse of a right-angled triangle has its endpoints at the points $(1, 3)$ and $(-4, 1)$. Find the equations of the legs (perpendicular sides) of the triangle.

What does the equation $\sqrt{(x - 2)^2 + (y + 1)^2} + \sqrt{(x + 2)^2 + (y + 4)^2} = 5$ represent?

Vedclass Test Series

Exam Paper Generator

Online Exam Module