The locus of a point which divides the join o | vedclass

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

Question

(B) Let the variable point on the circle be $B(x_1, y_1)$. Since $B$ lies on the circle ${x^2} + {y^2} = 4$,we have ${x_1}^2 + {y_1}^2 = 4$.Let the po

Vedclass · Accepted Answer

$25({x^2} + {y^2}) + 20(x - y) - 28 = 0$

Vedclass · Answer

(B) Let the variable point on the circle be $B(x_1, y_1)$. Since $B$ lies on the circle ${x^2} + {y^2} = 4$,we have ${x_1}^2 + {y_1}^2 = 4$.Let the point dividing the segment $AB$ in the ratio $3 : 2$ be $P(h, k)$.Using the section formula,we have:$h = \frac{3x_1 + 2(-1)}{3 + 2} = \frac{3x_1 - 2}{5} \implies x_1 = \frac{5h + 2}{3}$$k = \frac{3y_1 + 2(1)}{3 + 2} = \frac{3y_1 + 2}{5} \implies y_1 = \frac{5k - 2}{3}$Substituting these values into the circle equation ${x_1}^2 + {y_1}^2 = 4$:$(\frac{5h + 2}{3})^2 + (\frac{5k - 2}{3})^2 = 4$$\frac{25h^2 + 20h + 4}{9} + \frac{25k^2 - 20k + 4}{9} = 4$$25(h^2 + k^2) + 20(h - k) + 8 = 36$$25(h^2 + k^2) + 20(h - k) - 28 = 0$Replacing $(h, k)$ with $(x, y)$,the locus is $25(x^2 + y^2) + 20(x - y) - 28 = 0$.

The locus of a point which divides the join of $A(-1, 1)$ and a variable point $B$ on the circle ${x^2} + {y^2} = 4$ in the ratio $3 : 2$ is:

Similar Questions

The locus of a point which moves in such a way that its distance from $(0,0)$ is three times its distance from the $x$-axis is given by:

If a chord is drawn from the origin to the circle $(x - 1)^2 + y^2 = 1$,then the equation of the locus of the midpoint of this chord is:

For any value of $\theta$,if the straight lines $x \sin \theta + (1 - \cos \theta) y = a \sin \theta$ and $x \sin \theta - (1 + \cos \theta) y + a \sin \theta = 0$ intersect at $P(\theta)$,then the locus of $P(\theta)$ is a

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

If the locus of the point,whose distances from the point $(2,1)$ and $(1,3)$ are in the ratio $5:4$,is $ax^2+by^2+cxy+dx+ey+170=0$,then the value of $a^2+2b+3c+4d+e$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module