Two line segments AB and CD are constrained t | vedclass

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

Question

(B) Let the equation of the circle passing through $A, B, C, D$ be $x^2+y^2+2gx+2fy+c=0$.The polar coordinates of the centre $(-g, -f)$ are given by $

Vedclass · Accepted Answer

$r^2 \cos 2	heta = \frac{a^2-b^2}{4}$

Vedclass · Answer

(B) Let the equation of the circle passing through $A, B, C, D$ be $x^2+y^2+2gx+2fy+c=0$.The polar coordinates of the centre $(-g, -f)$ are given by $-g = r \cos 	heta$ and $-f = r \sin 	heta$.$AB$ is the length of the intercept on the $X$-axis,so $a = 2\sqrt{g^2-c} \implies \frac{a^2}{4} = g^2-c$.$CD$ is the length of the intercept on the $Y$-axis,so $b = 2\sqrt{f^2-c} \implies \frac{b^2}{4} = f^2-c$.Subtracting the two equations: $g^2-f^2 = \frac{a^2-b^2}{4}$.Substituting $g = -r \cos 	heta$ and $f = -r \sin 	heta$:$(-r \cos 	heta)^2 - (-r \sin 	heta)^2 = \frac{a^2-b^2}{4}$$r^2(\cos^2 	heta - \sin^2 	heta) = \frac{a^2-b^2}{4}$$r^2 \cos 2	heta = \frac{a^2-b^2}{4}$.

Two line segments $AB$ and $CD$ are constrained to move along the $X$ and $Y$-axes,respectively,in such a way that the points $A, B, C, D$ are concyclic. If $AB = a$ and $CD = b$,then the locus of the centre of the circle passing through $A, B, C, D$ in polar coordinates is

Similar Questions

The angle between the tangents drawn from a point $P$ to the circle $x^2 + y^2 + 4x - 2y - 4 = 0$ is $60^{\circ}$. The locus of $P$ is:

The locus of the midpoints of the chords of the circle $x^{2}+y^{2}=4$ which subtend a right angle at the origin is

$A$ point $P$ moves such that the distance from $(0,2)$ to $P$ is $\frac{1}{\sqrt{2}}$ times the distance of $P$ from $(-1,0)$. Then the locus of the point is

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:

The locus of the image of the point $(2, 3)$ in the line $(2x - 3y + 4) + k(x - 2y + 3) = 0, k \in R$ is a:

Vedclass Test Series

Exam Paper Generator

Online Exam Module