A point moves such that the sum of the square | vedclass

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

Question

(B) Let $P(h, k)$ be any point on the locus.Given that the sum of the squares of its distances from $A(1, 2)$ and $B(-2, 1)$ is $6$.$(PA)^2 + (PB)^2 =

Vedclass · Accepted Answer

a circle with centre $(-\frac{1}{2}, \frac{3}{2})$ and radius $\frac{1}{\sqrt{2}}$

Vedclass · Answer

(B) Let $P(h, k)$ be any point on the locus.Given that the sum of the squares of its distances from $A(1, 2)$ and $B(-2, 1)$ is $6$.$(PA)^2 + (PB)^2 = 6$$(h - 1)^2 + (k - 2)^2 + (h + 2)^2 + (k - 1)^2 = 6$$(h^2 - 2h + 1) + (k^2 - 4k + 4) + (h^2 + 4h + 4) + (k^2 - 2k + 1) = 6$$2h^2 + 2k^2 + 2h - 6k + 10 = 6$$2h^2 + 2k^2 + 2h - 6k + 4 = 0$$h^2 + k^2 + h - 3k + 2 = 0$Replacing $(h, k)$ with $(x, y)$, the locus is $x^2 + y^2 + x - 3y + 2 = 0$.This is the equation of a circle.The centre is $(-\frac{1}{2}, \frac{3}{2})$ and the radius is $\sqrt{(-\frac{1}{2})^2 + (\frac{3}{2})^2 - 2} = \sqrt{\frac{1}{4} + \frac{9}{4} - 2} = \sqrt{\frac{10}{4} - 2} = \sqrt{\frac{5}{2} - 2} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.

$A$ point moves such that the sum of the squares of its distances from the points $(1, 2)$ and $(-2, 1)$ is always $6$. Then, its locus is

Similar Questions

Find the locus of a point such that its distance from the point $(0, -1)$ is twice its distance from the line $3x + 4y + 1 = 0$.

If the sum of the distances of a point from the origin and the line $x = 2$ is $4$,then its locus is

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 to the circle $x^2 + y^2 = a^2$ with slopes $\alpha$ and $\beta$ intersect at point $P$,and $\cot \alpha + \cot \beta = 0$,then the locus of $P$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module