CircleHardBloom L3
Question
A circle of constant radius ' $a$ ' passes through origin ' $O$ ' and cuts the axes of coordinates in points $P$ and $Q$, then the equation of the locus of the foot of perpendicular from $O$ to $PQ$ is:
Options
A.$\left( x^{2} + y^{2} \right)\left( \frac{1}{x^{2}} + \frac{1}{y^{2}} \right) = 4a^{2}$
B.$\left( x^{2} + y^{2} \right)^{2}\left( \frac{1}{x^{2}} + \frac{1}{y^{2}} \right) = a^{2}$
C.$\left( x^{2} + y^{2} \right)^{2}\left( \frac{1}{x^{2}} + \frac{1}{y^{2}} \right) = 4a^{2}$
D.$\left( x^{2} + y^{2} \right)\left( \frac{1}{x^{2}} + \frac{1}{y^{2}} \right) = a^{2}$
Solution
[Key Concept: A radius-$a$ circle through $O$ cutting the axes at $A,B$: eliminate to get the locus of the centroid/foot.]
Equation of $PQ$ is
$$hx + ky = h^{2} + k^{2}\left( 0,\frac{h^{2} + k^{2}}{k} \right) $$
$$\begin{matrix} & PQ^{2} = 4a^{2} \\ & \left( h^{2} + k^{2} \right)^{2}\left( \frac{1}{h^{2}} + \frac{1}{k^{2}} \right) = 4a^{2} \end{matrix}$$
Create a free account to view solution
View Solution FreeMore Circle Questions
The circle x2 + y2 − 4x − 4y + 4 = 0 is-...The number of tangents that can be drawn from the point (8, 6) to the circle x2 + y2 _ 100 = 0 is...M lies on the curveThe length of the diameter of the circle x2 + y2 − 4x − 6y + 4 = 0 is -...The equation of a line inclined at an angle $\frac{\pi}{4}$ to the axis $X$, such that the two circles $x^{2} + y^{2} = ...If the latus of an hyperbola be 8 and eccentricity be then the equation of the hyper bola can be -...