If a circle of constant radius 3k passes thro | vedclass

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

Question

(A) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$.Since the circle passes through the origin $(0, 0)$,$A(a, 0)$,and $B(0, b)$,the equatio

Vedclass · Accepted Answer

$x^2 + y^2 = (2k)^2$

Vedclass · Answer

(A) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$.Since the circle passes through the origin $(0, 0)$,$A(a, 0)$,and $B(0, b)$,the equation of the circle is $x^2 + y^2 - ax - by = 0$.The radius of this circle is given by $\sqrt{(\frac{a}{2})^2 + (\frac{b}{2})^2} = 3k$.Squaring both sides,we get $\frac{a^2}{4} + \frac{b^2}{4} = 9k^2$,which implies $a^2 + b^2 = 36k^2$.Let $(x, y)$ be the centroid of $	riangle OAB$. Then $x = \frac{0+a+0}{3} = \frac{a}{3}$ and $y = \frac{0+0+b}{3} = \frac{b}{3}$.Thus,$a = 3x$ and $b = 3y$.Substituting these into the equation $a^2 + b^2 = 36k^2$,we get $(3x)^2 + (3y)^2 = 36k^2$.$9x^2 + 9y^2 = 36k^2$.Dividing by $9$,we get $x^2 + y^2 = 4k^2 = (2k)^2$.

If a circle of constant radius $3k$ passes through the origin $O$ and meets the coordinate axes at $A$ and $B$,then the locus of the centroid of the triangle $OAB$ is

Similar Questions

Find the locus of a point from which the lengths of the tangents drawn to the two circles $x^2 + y^2 - 5x - 3 = 0$ and $3x^2 + 3y^2 + 2x + 4y - 6 = 0$ are equal.

If $A(1, 1)$,$B(-1, 1)$,and $C(-1, -1)$ are three points and a point $P(x, y)$ moves such that $PA^2 = PB^2 + PC^2$,then the equation of the locus of $P$ is:

$A$ variable straight line $AB$ divides the circumference of the circle $x^2 + y^2 = 25$ in the ratio $1 : 2$. If a tangent $CD$ is drawn to the smaller arc parallel to $AB$,such that $ABCD$ is a rectangle,then the locus of $C$ and $D$ is:

If 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 $P$ is:

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 $|a|=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module