Question

The pedal equation of the circle $x^2+y^2=4$ with regard to the point (2,0) is
[a] $r^2=4$
[b] $4p=r^2$
[c] $p^2=4r$
[d] $p=4r^2$

Answer 1

Hint: Use the property that the tangent of a circle and the radius are perpendicular to each other at the point of contact. Use the property that angle in a semicircle is a right angle. Hence prove that the angles $\mathrm{\angle }\text{ABD}$ and $\mathrm{\angle }\text{BCA}$ are equal. Use trigonometry to get a relation between AB, AD and BC. Substitute AD = p, AB = r and BC = 4 to get the required pedal equation.

Complete step-by-step answer:

Pedal equation: Definition: Pedal equation of a curve C and given fixed point O is the relation between r and p, where r is the distance if point O to a point on the curve C and p is the perpendicular distance of O from the tangent line drawn at that point. The point O is called the pedal point and the values r and p are known as the pedal coordinates of the curve C w.r.t the pedal point.
Here we have, the pedal point is A (2,0), and B is any point on the locus. BD is the tangent at point B, and AD is the perpendicular from A on BD. C is the diametrically opposite point of B.
Hence, we have AD = p and AB = r.
Since BC is a diameter, we have $\mathrm{\angle }\text{BAC=90}{}^{\circ }$
Let $\mathrm{\angle }\text{BCA=}\theta$.
Hence we have
$\mathrm{\angle }\text{CBA=90}{}^{\circ }\text{-}\theta$
Since BD is a tangent and BC is a diameter, we have $\mathrm{\angle }\text{CBD=90}{}^{\circ }$
Hence, we have $\mathrm{\angle }\text{ABD=}\theta$
In triangle ABD, we have $\sin \theta ={\displaystyle \frac{\text{AD}}{\text{AB}}}\text{ (i)}$
In triangle ABC, we have $\sin \theta ={\displaystyle \frac{\text{AB}}{\text{BC}}}\text{ (ii)}$
From (i) and (ii), we get
${\displaystyle \frac{\text{AD}}{\text{AB}}}={\displaystyle \frac{\text{AB}}{\text{BC}}}$
Since for circle with equation $x^2+y^2=a^2$ radius = a, we have BC = 4.
Put AD = p, AB = r and BC = 4, we get
$\begin{array}{rl}& {\displaystyle \frac{p}{r}}={\displaystyle \frac{r}{4}}\\ & \Rightarrow 4p=r^2\end{array}$
which is the required pedal equation of the circle.
Hence option [b] is correct.

Note: In the pedal equation, we denote distance of any point on the locus from the pedal point by r and the distance of the pedal point from the tangent at the point on the locus by p. We then find the relation between p and r. This relation is called the pedal equation of the curve.
Alternative Solution:
Pedal equation of a circle with respect to any point on the circumference of the circle is given by $pd=r^2$, where d is the diameter of the circle.
Here d = 4.
Hence the pedal equation is $4p=r^2$ .