Find the equation of tangent through the point (3,4) on the circle $x^2+y^2 = 9$

Question

radius = 3
Let $y-y_1 = m(x-x_1)$ be the equation of tangent. Since, tangent passes through (3,4)
$or,\text{ }y - 4 = m(x - 3)$
$or,\text{ }y - 4 = mx - 3m$
$or,\text{ }mx +y - 3m + 4 = 0$
Thus, center of given circle is $(0,0)$. So, the perpendicular distance from the center to the tangent.
$or,\text{ }p = \big | \frac{ax_1 + by_1+c}{\sqrt{a^2+b^2}}\big |$
$or,\text{ }p = \big | \frac{a.0 + b.0 +4-3m}{\sqrt{{m}^2+1^2}}\big |$
$or,\text{ }p = \big | \frac{4-3m}{\sqrt{{m}^2+1^2}}\big |$
$or,\text{ }p^2 = \frac{(4-3m)^2}{{{m}^2+1^2}}$
As perpendicular distance from tangent to center = radius of circle,
$or,\text{ }9m^2 + 9 = 16 + 9m^2 - 24m$
$or,\text{ }m = \frac{7}{24}$
I wonder why one solution only? and Solve this further

Answer 1

Let $(a,b)$ be a point on the circle $x^2+y^2 = 9$. Then the line $a(x-a)+b(y-b) = 0$ is line tangent to the circle that passes through point $(a,b)$. We also want point $(3,4)$ to be on the line $a(x-a)+b(y-b) = 0$ so we get system

\begin{align} a(3-a)+b(4-b) &= 0\\ a^2+b^2 &= 9 \end{align}

If we expand the first equation we get $3a+4b-(a^2+b^2) = 0$ and substituting $a^2+b^2 = 9$, the system becomes

\begin{align} 3a+4b &= 9\\ a^2+b^2 &= 9 \end{align}

From the first equation we get $a = 3-\frac 43 b$, which you can substitute into the second equation and get two solutions $a=3,b=0$ and $a=-\frac{21}{25}, b=\frac{72}{25}$.

Finally, substitute that back into tangent line equation $a(x-a)+b(y-b) = 0$, first solution gives the line $3(x-3) + 0(y-0) = 0$, i.e. $x = 3$ and the second solution gives the line $-\frac{21}{25}(x+\frac{21}{25})+\frac{72}{25}(y-\frac{72}{25}) = 0$, i.e. $y = \frac 7{24}x+\frac{25}8.$

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Your approach is correct and gives the tangent line $y = \frac 7{24}x+\frac{25}8$, but not the line $x = 3$. This is because you assumed that tangent line is of the form $y-y_1 = m(x-x_1)$, but $x = 3$ is not of that form, so you only got one line. If you want to finish the exercise your way, you can repeat the process but assume that you have tangent line of the form $m(y-y_1) = x-x_1$. You should get $m = 0$.

Answer 2

I am actually busy right now but I will expand my answer later of you wish.

Use the $SS_1=T^2$ method and then factorise the resultant expression. You'll get both the tangents.

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In this answer $S$ denotes the equation of circle, $S_1$ denotes the numerical value we get after putting the coordinate given in the equation of the circle and $T$ denotes the equation of chord of contact.

Proof click here

Answer 3

The fact that the $m^2$ term cancels from your equation indicates that $m=\infty$ is a valid solution. This value corresponds to the vertical tangent.

Answer 4

HINT

Calculus-based answer. I would start with noticing that \begin{align*} x^{2} + y^{2} = 9 \Rightarrow 2x + 2yy' = 0 \Rightarrow y' = -\frac{x}{y} \end{align*}

Let $(x_{0},y_{0})\in \partial B_{3}(0)$. If $y_{0} = 0$, then $(x_{0},y_{0}) = (3,0)$ is a solution.

If $y_{0}\neq 0$, one gets the following system of equations to solve: \begin{align*} \begin{cases} x^{2}_{0} + y^{2}_{0} = 9\\\\ y_{0} - 4 = -\dfrac{x_{0}}{y_{0}}(x_{0} - 3) \end{cases} & \Longleftrightarrow \begin{cases} x^{2}_{0} + y^{2}_{0} = 9\\\\ x^{2}_{0} + y^{2}_{0} - 3x_{0} - 4y_{0} = 0 \end{cases} \Longleftrightarrow \begin{cases} x^{2}_{0} + y^{2}_{0} = 9\\\\ 3x_{0} + 4y_{0} = 9 \end{cases} \end{align*}

Can you take it from here?