The equation of the locus of a point (x, y) w | vedclass

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

Question

(A) Let the point be $P(x, y)$.Given that the distance of $P$ from $(1, 4)$ is $5$,we have: $\sqrt{(x-1)^2 + (y-4)^2} = 5 \implies (x-1)^2 + (y-4)^2 =

Vedclass · Accepted Answer

$9x^2 + 12xy + 4y^2 - 30x - 108y + 222 = 0$

Vedclass · Answer

(A) Let the point be $P(x, y)$.Given that the distance of $P$ from $(1, 4)$ is $5$,we have: $\sqrt{(x-1)^2 + (y-4)^2} = 5 \implies (x-1)^2 + (y-4)^2 = 25 \implies x^2 - 2x + 1 + y^2 - 8y + 16 = 25 \implies x^2 + y^2 - 2x - 8y - 8 = 0$.Also,the distance of $P$ from the line $2x + 3y - 1 = 0$ is $5$,so: $\frac{|2x + 3y - 1|}{\sqrt{2^2 + 3^2}} = 5 \implies |2x + 3y - 1| = 5\sqrt{13}$.Squaring both sides: $(2x + 3y - 1)^2 = 25 	imes 13 = 325$.$4x^2 + 9y^2 + 1 + 12xy - 4x - 6y = 325 \implies 4x^2 + 12xy + 9y^2 - 4x - 6y - 324 = 0$.Since the locus must satisfy both conditions,we look for the intersection of these two loci. However,the question asks for the equation of the locus of a point that satisfies both conditions simultaneously. This implies the point must lie on the intersection of the circle and the two lines parallel to the given line. The question is likely asking for the locus of points equidistant from the point and the line,but given the phrasing,it implies the intersection points. Given the options,we check for the expansion of the intersection. The correct locus satisfying the distance condition from the line is $4x^2 + 12xy + 9y^2 - 4x - 6y - 324 = 0$.

The equation of the locus of a point $(x, y)$ which is at a distance of $5$ units from a fixed point $(1, 4)$ and also at a distance of $5$ units from a fixed line $2x + 3y - 1 = 0$ is:

Similar Questions

$A$ circle passes through the point $(3, 4)$ and cuts the circle $x^2 + y^2 = a^2$ orthogonally. The locus of its centre is a straight line. If the distance of this straight line from the origin is $25$,then $a^2$ is equal to:

The locus of the center of the circle which cuts the circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ orthogonally is

The locus of a point $P$ which divides the line joining $(1, 0)$ and $(2\cos \theta, 2\sin \theta)$ internally in the ratio $2 : 3$ for all $\theta$ is a

The set of all points that are at a distance of at least $2$ units from $(-3, 0)$ is

The locus of the centre of a circle passing through $(a, b)$ and cutting orthogonally to the circle $x^2 + y^2 = p^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module