A circle of constant radius a passes through | vedclass

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

Question

(C) Let the foot of the perpendicular from the origin $O(0,0)$ to the line $PQ$ be $R(h, k)$.Since $OR \perp PQ$,the slope of $OR$ is $m_1 = \frac{k}{

Vedclass · Accepted Answer

$(x^2 + y^2)^2 \left( \frac{1}{x^2} + \frac{1}{y^2} \right) = 4a^2$

Vedclass · Answer

(C) Let the foot of the perpendicular from the origin $O(0,0)$ to the line $PQ$ be $R(h, k)$.Since $OR \perp PQ$,the slope of $OR$ is $m_1 = \frac{k}{h}$.The slope of line $PQ$ is $m_2 = -\frac{h}{k}$.The equation of line $PQ$ passing through $(h, k)$ is $y - k = -\frac{h}{k}(x - h)$,which simplifies to $hx + ky = h^2 + k^2$.The intercepts of this line on the axes are $Q\left( \frac{h^2 + k^2}{h}, 0 ight)$ and $P\left( 0, \frac{h^2 + k^2}{k} ight)$.Since $O, P, Q$ lie on a circle of radius $a$,the diameter of the circle is the hypotenuse $PQ$ of the right-angled triangle $	riangle OPQ$.The length $PQ = \sqrt{\left( \frac{h^2 + k^2}{h} ight)^2 + \left( \frac{h^2 + k^2}{k} ight)^2} = 2a$.Squaring both sides: $(h^2 + k^2)^2 \left( \frac{1}{h^2} + \frac{1}{k^2} ight) = 4a^2$.Replacing $(h, k)$ with $(x, y)$,the locus is $(x^2 + y^2)^2 \left( \frac{1}{x^2} + \frac{1}{y^2} ight) = 4a^2$.

$A$ circle of constant radius $a$ passes through the origin $O$ and cuts the coordinate axes at points $P$ and $Q$. The equation of the locus of the foot of the perpendicular from $O$ to $PQ$ 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:

$A(a, 0)$ is a fixed point and $\theta$ is a parameter such that $0 < \theta < 2 \pi$. If $P(a \cos \theta, a \sin \theta)$ is a point on the circle $x^2+y^2=a^2$ and $Q(b \sin \theta, -b \cos \theta)$ is a point on the circle $x^2+y^2=b^2$,then the locus of the centroid of the triangle $APQ$ is

The equation of the locus of a point which moves so as to be at equal distances from the point $(a, 0)$ and the $y$-axis 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 $|a|=$

Let $A = (a, 0)$ and $B = (-a, 0)$ be two fixed points. For $a \in (-\infty, 0)$,point $P(x, y)$ moves in the plane such that $PA = nPB$ $(n \neq 0, n \neq 1)$. If $0 < n < 1$,then which of the following is true regarding the locus of $P$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module