How to solve this problem without letting coordinates as $(a \cos\theta , a\sin\theta)$

Question

Q. Tangent to the circle $x^2 + y^2 = a^2$ at any point on it in the first quadrant makes intercepts $OA$ and $OB$ on $x$ and $y$ axes respectively, $O$ being the centre of the circle. Find the minimum value of $OA + OB.$

The solution given everywhere includes the following solution :-

Let $P(a\cos\theta,a\sin\theta)$ be an arbitrary point on the circle $x^2 + y^2 = a^2$. If $P$ lies in the first quadrant, then $0<\theta<\frac{\pi}{2}$ .The equation of tangent to $x^2 + y^2 = a^2$ at $P(a\cos\theta,a\sin\theta)$ is $x\cos\theta + y\sin\theta=a$

$$ OA = a \sec\theta \quad \text{and} \quad OB = a \csc\theta $$

Let $S=OA + OB$. Then ,

$$ S=a(\sec\theta + \csc\ \theta) $$

$$ \frac{dS}{d\theta}=a(\sec\theta\cdot\tan\theta - \csc \ \theta \cdot\cot\theta) $$

and $$ \frac{d^2S}{d\theta^2} = a( \sec^3\theta + \sec\theta\cdot\tan^2\theta + \csc^3 \theta + \csc \theta\cdot\cot^2\theta \ ) $$

For maximum or minimum values of $S$ , we must have $$ \frac{dS}{d\theta}=0 $$

$$ a(\sec\theta\cdot\tan\theta - \csc \ \theta \cdot\cot\theta) = 0 $$

$$ \tan^3\theta = 1 $$

$$ \theta = \frac{\pi}{4} $$

At $\theta = \frac{\pi}{4}$ , we obtain

$$ \frac{d^2S}{d\theta^2} = a( \ 2\sqrt2 + \sqrt2 + 2\sqrt2 + \sqrt2 \ ) $$

$$ = 6\sqrt2a > 0 $$

Hence , $S$ is minimum at $\theta = \frac{\pi}{4}$ and the minimum value of $S$ is given by

$$ S = a\cdot(\sec\frac{\pi}{4}+ \csc \frac{\pi}{4}) = 2\sqrt{2}\cdot a $$

So my question is that is there any other way to solve this question other than letting the points as $(a\cos\theta,a\sin\theta)$ ?

Answer 1

You can use that the equation of the tangent line at the point $$P(x_0,y_0)$$ situated on the circle is given by $$xx_0+yy_0=a^2$$

Answer 2

Let angle $\angle{POA}$ be $\theta$ then $OA=asec{\theta}$ and $OB=acosec{\theta}$.

The remaining follows....

Answer 3

If the point has coordinates $(p,q)$, then the tangent line at it has equation $$ y-q=-\frac{p}{q}(x-p) $$ that can be rewritten as $qy-q^2+px-p^2=0$. Since $p^2+q^2=a^2$, we get $$ px+qy=a^2 $$ If $x=0$, we get $y=a^2/q$; if $y=0$ we get $x=a^2/p$. The function you have to minimize is thus $$ f(p,q)=\frac{a^2}{p}+\frac{a^2}{q},\qquad p^2+q^2=a^2,\quad p,q>0 $$ By letting $\xi=p/a$, the function to minimize becomes $$ g(\xi)=\frac{1}{\xi}+\frac{1}{\sqrt{1-\xi^2}},\qquad 0<\xi<1 $$ The derivative is $$ g'(\xi)=-\frac{1}{\xi^2}+\frac{\xi}{\sqrt{(1-\xi^2)^3}} $$ that vanishes where $$ \xi^6-(1-\xi^2)^3=0 $$ that can be factored as $(2\xi^2-1)(\xi^4+\xi^2(1-\xi^2)+(1-\xi^2)^2)=0$.

This only happens for $\xi^2=1/2$. So $p=q=a/\sqrt{2}$ and $$ f(a/\sqrt{2},a/\sqrt{2})=2a\sqrt{2} $$

Answer 4

The situation is symmetric with respect to $A$ and $B$, so we can expect that something interesting occurs when $OA=OB$, from which a quick calculation yields $OA+OB=2\sqrt2a$.

In more detail, the line with the given intercepts has the equation $$\frac x{x_A}+\frac y{y_B}=1.$$ This line is tangent to the circle when its distance from the origin is equal to $a$. Using the point-line distance formula, this constraint can be expressed as $${1\over\sqrt{1/x_A^2+1/x_B^2}} = a,$$ or $${x_A y_B \over \sqrt{x_A^2+y_B^2}} = a.\tag 1$$ You can now solve this constrained minimization problem via Lagrange multipliers, or proceed directly by solving (1) for either $x_A$ or $y_B$. Doing the latter, we then have to minimize $$x_A+y_B = x_A+{ax_A\over\sqrt{x_A^2+a^2}}$$ and a straightforward, though tedious, computation reveals a minimum at $x_A=a\sqrt2$.

A simpler approach is to make the geometric observation that $OA+OB$ is proportional to $\sec\theta+\csc\theta$, where $\theta$ is the angle between the $x$-axis and the radius at the point of tangency. Both of these trigonometric functions are positive on the interval $(0,\pi/2)$ and their sum is symmetric about $\theta=\pi/4$, so that must be a local minimum.