Tangent Lines To A Circle Through A Point

Alright, guys! Let's dive into a fun math problem: finding the tangent lines to a circle that pass through a specific point. We'll break it down step by step so it’s super easy to follow. Our circle is defined by the equation $L = x^2 + y^2 - 4x - 2y - 11 = 0$, and we want to find the tangent lines that go through the point $P(6, 5)$. Ready? Let’s go!

Understanding the Problem

Before we jump into calculations, let’s make sure we understand what we’re trying to do. A tangent line to a circle is a line that touches the circle at exactly one point. We have a circle and a point outside (or possibly on) the circle, and we need to find the equations of the lines that touch the circle only once and also pass through our given point. Visualizing this can really help, so maybe sketch a quick diagram!

Our circle equation is $L = x^2 + y^2 - 4x - 2y - 11 = 0$. We can rewrite this in the standard form $(x - a)^2 + (y - b)^2 = r^2$ to find the center and radius, which will be very useful. The point $P(6, 5)$ is the point through which our tangent lines must pass. This means that when we find the equations of the tangent lines, plugging in $x = 6$ and $y = 5$ should satisfy those equations.

Step 1: Finding the Circle's Center and Radius

First, complete the square to rewrite the circle equation in the standard form:

$x^2 - 4x + y^2 - 2y - 11 = 0$

To complete the square for the $x$ terms, we need to add and subtract $(4/2)^2 = 4$. For the $y$ terms, we add and subtract $(2/2)^2 = 1$.

$(x^2 - 4x + 4) + (y^2 - 2y + 1) - 11 - 4 - 1 = 0$

$(x - 2)^2 + (y - 1)^2 = 16$

Now we can easily see that the center of the circle is $(2, 1)$ and the radius is $\sqrt{16} = 4$.

Step 2: General Equation of a Line Through $P(6, 5)$

We know that the tangent line passes through the point $P(6, 5)$. The general equation of a line can be written as $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept. Since the line passes through $(6, 5)$, we can write:

$5 = 6m + c$

So, $c = 5 - 6m$. Now we can rewrite the equation of the line as:

$y = mx + (5 - 6m)$

$y = mx - 6m + 5$

Step 3: Using the Distance Formula

For a line to be tangent to the circle, the distance from the center of the circle to the line must be equal to the radius of the circle. The distance $d$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by:

$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

First, rewrite our line equation $y = mx - 6m + 5$ in the form $Ax + By + C = 0$:

$mx - y - 6m + 5 = 0$

Here, $A = m$, $B = -1$, and $C = -6m + 5$. The center of our circle is $(2, 1)$, so $x_0 = 2$ and $y_0 = 1$. The radius is $4$. Plugging these values into the distance formula:

$4 = \frac{|m(2) - 1 - 6m + 5|}{\sqrt{m^2 + (-1)^2}}$

$4 = \frac{|2m - 1 - 6m + 5|}{\sqrt{m^2 + 1}}$

$4 = \frac{|-4m + 4|}{\sqrt{m^2 + 1}}$

Step 4: Solving for $m$

Now we need to solve for $m$. Square both sides of the equation to get rid of the square root and the absolute value:

$16 = \frac{(-4m + 4)^2}{m^2 + 1}$

$16(m^2 + 1) = (16m^2 - 32m + 16)$

$16m^2 + 16 = 16m^2 - 32m + 16$

Notice that $16m^2$ cancels out on both sides, leaving us with:

$16 = -32m + 16$

$32m = 0$

$m = 0$

This gives us one value for $m$. But we need to consider the case where $-4m + 4$ could be negative, so let's go back to the equation $4 = \frac{|-4m + 4|}{\sqrt{m^2 + 1}}$ and rewrite it as:

$4\sqrt{m^2 + 1} = |-4m + 4|$

$4\sqrt{m^2 + 1} = |4 - 4m|$

Now consider the case where $4 - 4m < 0$, so $|4 - 4m| = -(4 - 4m) = 4m - 4$:

$4\sqrt{m^2 + 1} = 4m - 4$

$\sqrt{m^2 + 1} = m - 1$

Square both sides:

$m^2 + 1 = m^2 - 2m + 1$

$1 = -2m + 1$

$2m = 0$

$m = 0$

That doesn't give us anything new. Let's consider the correct squaring: $16(m^2+1) = (-4m+4)^2$ $16m^2 + 16 = 16m^2 -32m + 16$ $0 = -32m$ $m = 0$ Since we squared the equation, we have to consider that the absolute value could be either positive or negative. Let's try that now: $4 = \frac{|-4m + 4|}{\sqrt{m^2 + 1}}$ $4\sqrt{m^2 + 1} = |-4m+4|$ So, $-4m+4 = 4\sqrt{m^2+1}$ or $-4m+4 = -4\sqrt{m^2+1}$. If $-4m+4 = -4\sqrt{m^2+1}$, then $4-4m < 0$, so $4 < 4m$, and $1 < m$.

Square both sides: $(1-m)^2 = m^2+1$, $1-2m+m^2=m^2+1$, $-2m=0$, so $m=0$. But that contradicts $1 < m$.

If $-4m+4 = 4\sqrt{m^2+1}$, then $1-m = \sqrt{m^2+1}$. So $(1-m)^2 = m^2+1$, $1-2m+m^2 = m^2+1$, and $-2m=0$, so $m=0$.

However, we made an error earlier. When we squared both sides, we should have had: $16(m^2 + 1) = (-4m + 4)^2$ $16m^2 + 16 = 16m^2 - 32m + 16$ $0 = -32m$ $m = 0$

It seems we only get $m = 0$. Let's reconsider the problem.

Another Approach: Using the Quadratic Formula

The equation of the line is $y - 5 = m(x - 6)$, or $y = mx - 6m + 5$. Substitute this into the circle equation $x^2 + y^2 - 4x - 2y - 11 = 0$: $x^2 + (mx - 6m + 5)^2 - 4x - 2(mx - 6m + 5) - 11 = 0$ $x^2 + m^2x^2 - 12m^2x + 36m^2 + 10mx - 60m + 25 - 4x - 2mx + 12m - 10 - 11 = 0$ $(1 + m^2)x^2 + (-12m^2 + 8m - 4)x + (36m^2 - 48m + 4) = 0$ For the line to be tangent, the discriminant must be 0: $D = b^2 - 4ac = (-12m^2 + 8m - 4)^2 - 4(1 + m^2)(36m^2 - 48m + 4) = 0$ $(144m^4 - 192m^3 + 160m^2 - 64m + 16) - 4(36m^2 - 48m + 4 + 36m^4 - 48m^3 + 4m^2) = 0$ $144m^4 - 192m^3 + 160m^2 - 64m + 16 - 144m^4 + 192m^3 - 160m^2 + 192m - 16 = 0$ $128m = 0$ $m = 0$ Once again, we only find $m = 0$. This implies we made a mistake.

Corrected Approach $y-5 = m(x-6)$, so $y = m(x-6) + 5$. Substitute this into $x^2 + y^2 - 4x - 2y - 11 = 0$. $x^2 + (m(x-6) + 5)^2 - 4x - 2(m(x-6)+5) - 11 = 0$ $x^2 + (mx - 6m + 5)^2 - 4x - 2mx + 12m - 10 - 11 = 0$ $x^2 + m^2x^2 + 36m^2 + 25 - 12m^2x + 10mx - 60m - 4x - 2mx + 12m - 21 = 0$ $(1+m^2)x^2 + (-12m^2+8m-4)x + (36m^2 - 48m + 4) = 0$ We set the discriminant to zero: $(-12m^2+8m-4)^2 - 4(1+m^2)(36m^2-48m+4) = 0$ $144m^4 + 64m^2 + 16 - 192m^3 + 96m^2 - 64m - 4(36m^2 - 48m + 4 + 36m^4 - 48m^3 + 4m^2) = 0$ $144m^4 - 192m^3 + 160m^2 - 64m + 16 - 144m^4 + 192m^3 - 160m^2 + 192m - 16 = 0$ $128m = 0$ thus $m=0$.

If $m=0$, $y=5$. Subbing that into the original circle gives $x^2 + 25 - 4x - 10 - 11 = 0$, so $x^2 - 4x + 4 = 0$, $(x-2)^2 = 0$, $x=2$. Since the solution is $x=2$, we have one point, thus one tangent line.

Let's consider the alternative. Rewrite as $y = mx - 6m + 5$, and $x = (y-5+6m)/m$. Then $(\frac{y-5+6m}{m})^2 + y^2 - 4(\frac{y-5+6m}{m}) - 2y - 11 = 0$ $(y-5+6m)^2 + m^2y^2 - 4m(y-5+6m) - 2m^2y - 11m^2 = 0$ $y^2 + 25 + 36m^2 - 10y + 12my - 60m + m^2y^2 - 4my + 20m - 24m^2 - 2m^2y - 11m^2 = 0$ $(1+m^2-2m^2)y^2 + (-10+12m-4m)y + (25+36m^2-60m+20m-24m^2-11m^2) = 0$ $(1-m^2)y^2 + (8m-10)y + (25+m^2-40m) = 0$ Thus $(8m-10)^2 - 4(1-m^2)(25+m^2-40m) = 0$ $64m^2 - 160m + 100 - 4(25 + m^2 - 40m - 25m^2 - m^4 + 40m^3) = 0$ $64m^2 - 160m + 100 - 100 - 4m^2 + 160m + 100m^2 + 4m^4 - 160m^3 = 0$ $4m^4 - 160m^3 + 160m^2 = 0$ $m^2(4m^2 - 160m + 160) = 0$

$m=0$ or $4m^2 - 160m + 160 = 0$ $m^2 - 40m + 40 = 0$ $m = (40 \pm \sqrt{1600 - 160})/2 = 20 \pm \sqrt{400-40} = 20 \pm \sqrt{360} = 20 \pm 6\sqrt{10}$

Step 5: Finding the Tangent Lines

We found the slopes $m = 0$, $m = 20 + 6\sqrt{10}$, and $m = 20 - 6\sqrt{10}$. Now we can find the equations of the tangent lines using $y = mx - 6m + 5$.

For $m = 0$:

$y = 0*x - 6*0 + 5$

$y = 5$

For $m = 20 + 6\sqrt{10}$:

$y = (20 + 6\sqrt{10})x - 6(20 + 6\sqrt{10}) + 5$

$y = (20 + 6\sqrt{10})x - 120 - 36\sqrt{10} + 5$

$y = (20 + 6\sqrt{10})x - 115 - 36\sqrt{10}$

For $m = 20 - 6\sqrt{10}$:

$y = (20 - 6\sqrt{10})x - 6(20 - 6\sqrt{10}) + 5$

$y = (20 - 6\sqrt{10})x - 120 + 36\sqrt{10} + 5$

$y = (20 - 6\sqrt{10})x - 115 + 36\sqrt{10}$

Final Answer

The equations of the tangent lines are:

1. $y = 5$
2. $y = (20 + 6\sqrt{10})x - 115 - 36\sqrt{10}$
3. $y = (20 - 6\sqrt{10})x - 115 + 36\sqrt{10}$

So there you have it! Finding tangent lines can be a bit tricky, but breaking it down step by step makes it manageable. Hope this helps you guys out!