Find the equation of the tangent.

Question

Find the equation to the tangent to the circle $x^2 + y^2 = a^2$ which passes through the point $(b,0)$.

Considering $m$ as the slope of the tangent I get $y=m(x+b)$. If $(p,q)$ be the point of contact then $m=-q/p$(perpendicular). equating the two and solving leads me to a very complicated term, which perhaps, ain't the answer.Is there any better way to solve it?

Answer 1

If $(a\cos t, a \sin t)$ is the point of contact, tangent to $S=0$ is $S_1=0$.

$$\implies x \cos t+y\sin t=a$$

This passes through $(b,0)$ $$\implies \cos t=a/b$$

Clearly, there are $2$ $t$ values for this. You can find the locus of the tangents alternatively.

Answer 2

We will copute the equation of the tangent that has a contact for positive $y$ (as there are two tangents, symmetric w.r.t the $x$ axis). I also assume $b>0$ (if $b<0$, there must be absolute values in some places below).

If $O$ is the center of the circle, $M$ the point $(b,0)$ and $T$ the contact point with (yet unknown) coordinates $(x,y)$, then the $OTM$ is a right triangle, with right angle $\hat T$. You thus have

$$\cos \hat O=\frac ab=\frac xa$$

Hence $x=\dfrac{a^2}b$.

Now, $T$ is on the circle, hence $x^2+y^2=a^2$, and since we want $y>0$, you get

$$y=\sqrt{a^2-x^2}=a\sqrt{1-\frac{a^2}{b^2}}=\frac{a}{b}\sqrt{b^2-a^2}$$

Notice that you must have $b>a$, as expected since the point $M$ must be outside the circle, otherwise there is no tangent at all.

Now you have two points on a line, it should be straightforward to compute its equation.

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Another, easier way to do this.

A generic point $P(x,y)$ on the tangent line is such that $\vec {MP}$ is orthogonal to $\vec{OT}$, and

$$\vec{OT}=\left(\begin{matrix}a\cos\hat O\\ a\sin\hat O \end{matrix}\right)$$

Hence the equation of the tangent is $\vec {MP}\cdot\vec{OT}=0$ or, dividing by $a$:

$$(x-b)\cos \hat O+y\sin\hat O=0$$

With $\cos \hat O=\frac ab$ and $\sin \hat O=\pm\sqrt{1-\frac{a^2}{b^2}}$. Take the $+$ sign for a contact point with positive $y$.

Multiplying by $b$, the equation simplifies to

$$a(x-b)+\sqrt{b^2-a^2}y=0$$

or

$$ax+\sqrt{b^2-a^2}y=ab$$

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Check the equation with the two known points: for $y=0$, you get $x=b$ as expected, and for $x=\frac{a^2}{b}$,

$$\sqrt{b^2-a^2}y=ab-\frac{a^3}{b}=\frac ab(b^2-a^2)$$

Hence $y=\frac ab\sqrt{b^2-a^2}$.

Answer 3

Polar coordinates :

$x=a\cos \theta;$ $ y= a \sin \theta.$

Triangle $(0,0)$, point of tangency, $(b,0)$ is a right triangle.

$\tan \theta = \dfrac{\sqrt{b^2-a^2}}{a}.$

Let slope of tangent be $m:$

$m= - \dfrac{a}{\sqrt{b^2-a^2}}.$

Equation of one of the tangents is : $y = m(x-b).$

Equation of the second tangent is?