Line-Circle Intersection: Finding 'm' For Two Distinct Points

Hey math whizzes and problem solvers! Today, we're diving deep into a super cool geometry problem that involves lines and circles. Specifically, we're on a mission to find the set of values for 'm' that makes the line defined by the equation $y = mx + 1$ cut through the circle $(x - 7)^2 + (y - 5)^2 = 20$ at not just one, but two separate and distinct points. This isn't just some abstract math puzzle; understanding how lines interact with circles is foundational in many areas, from computer graphics and navigation systems to physics and engineering. So, grab your favorite thinking cap, maybe a snack, and let's get this done!

The Setup: What Are We Working With?

Alright, let's break down what we've got. We have a circle with the equation $(x - 7)^2 + (y - 5)^2 = 20$. This tells us a few key things right off the bat. The center of our circle is chilling at the coordinates $(7, 5)$, and its radius is the square root of 20, which simplifies to $2\sqrt{5}$. So, imagine a circle perfectly placed on the coordinate plane, centered at (7, 5). Now, we also have a line with the equation $y = mx + 1$. This line is a bit more dynamic because of the 'm'. 'm' represents the slope of the line. As 'm' changes, the line pivots around a specific point. What point is that? Well, if we set $x = 0$ in the line's equation, we get $y = m(0) + 1$, which means $y = 1$. So, no matter what the slope 'm' is, our line will always pass through the point $(0, 1)$. This is a crucial piece of information, guys!

The core question is: for which values of 'm' will this line $y = mx + 1$ actually cross our circle $(x - 7)^2 + (y - 5)^2 = 20$ twice? If it touches the circle at just one point, that's called being tangent. If it misses the circle entirely, there are no intersection points. We're specifically looking for the cases where there are two distinct intersection points. This means the line is a secant line to the circle. To figure this out mathematically, we need to find a way to combine these two equations and see how many solutions we get for 'x' (or 'y').

The Strategy: Substitution is Key!

So, how do we actually find these values of 'm'? The most common and effective strategy here is substitution. We have two equations, and we want to find points $(x, y)$ that satisfy both simultaneously. Since the line equation gives us $y$ in terms of $x$ (i.e., $y = mx + 1$), we can substitute this expression for $y$ into the circle's equation. This will give us a single equation with only one variable, $x$, and the parameter 'm' we're trying to solve for.

Let's do the substitution:

Start with the circle equation: $(x - 7)^2 + (y - 5)^2 = 20$

Now, substitute $y = mx + 1$ into it:

$(x - 7)^2 + ((mx + 1) - 5)^2 = 20$

Simplify the second term inside the parenthesis: $(mx + 1) - 5 = mx - 4$.

So, the equation becomes:

$(x - 7)^2 + (mx - 4)^2 = 20$

Now, we need to expand both of these squared terms. Remember the formulas $(a - b)^2 = a^2 - 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2$ (which is actually $(a+b)^2$ if we were using $mx+1$ directly, but since we simplified to $mx-4$ it's $(a-b)^2$).

Expanding $(x - 7)^2$: $x^2 - 14x + 49$

Expanding $(mx - 4)^2$: $(mx)^2 - 2(mx)(4) + 4^2 = m^2x^2 - 8mx + 16$

Now, put it all together back into our equation:

$(x^2 - 14x + 49) + (m^2x^2 - 8mx + 16) = 20$

Let's group the terms by powers of $x$. This is going to turn into a quadratic equation in terms of $x$. We want to collect all the $x^2$ terms, all the $x$ terms, and all the constant terms.

$(1 + m^2)x^2 + (-14 - 8m)x + (49 + 16) = 20$

Combine the constant terms: $49 + 16 = 65$.

$(1 + m^2)x^2 + (-14 - 8m)x + 65 = 20$

Finally, move the 20 to the left side to set the equation to zero, which is the standard form for a quadratic equation ($Ax^2 + Bx + C = 0$):

$(1 + m^2)x^2 + (-14 - 8m)x + (65 - 20) = 0$

$(1 + m^2)x^2 + (-14 - 8m)x + 45 = 0$

And there we have it! A quadratic equation in $x$. The coefficients are:

A = $1 + m^2$ B = $-14 - 8m$ C = $45$

This quadratic equation represents the $x$-coordinates of the intersection points. The number of solutions for $x$ tells us the number of intersection points. If we get two distinct real solutions for $x$, it means our line intersects the circle at two distinct points. If we get exactly one real solution, the line is tangent. If we get no real solutions (complex solutions), the line misses the circle.

The Condition for Two Distinct Points: The Discriminant

Alright guys, we've got our quadratic equation: $(1 + m^2)x^2 + (-14 - 8m)x + 45 = 0$. Now, how do we ensure there are two distinct points of intersection? This is where the discriminant comes into play. For any quadratic equation in the form $Ax^2 + Bx + C = 0$, the discriminant is given by the formula $\Delta = B^2 - 4AC$. The value of the discriminant tells us about the nature of the roots (solutions) of the quadratic equation:

• If $\Delta > 0$, there are two distinct real roots.
• If $\Delta = 0$, there is exactly one real root (a repeated root).
• If $\Delta < 0$, there are no real roots (two complex conjugate roots).

Since we want two distinct points of intersection, we need our quadratic equation in $x$ to have two distinct real roots. This means the discriminant must be greater than zero ($\Delta > 0$).

Let's plug in our coefficients A, B, and C into the discriminant formula:

A = $1 + m^2$ B = $-14 - 8m$ C = $45$

$\Delta = (-14 - 8m)^2 - 4(1 + m^2)(45)$

First, let's simplify $(-14 - 8m)^2$. We can factor out a -1: $(-1(14 + 8m))^2 = (14 + 8m)^2$. Now expand this:

$(14 + 8m)^2 = 14^2 + 2(14)(8m) + (8m)^2 = 196 + 224m + 64m^2$.

Next, let's simplify $4(1 + m^2)(45)$. Multiply the constants first: $4 \times 45 = 180$. So, we have $180(1 + m^2)$.

$180(1 + m^2) = 180 + 180m^2$.

Now, substitute these back into the discriminant equation:

$\Delta = (196 + 224m + 64m^2) - (180 + 180m^2)$

Remove the parentheses:

$\Delta = 196 + 224m + 64m^2 - 180 - 180m^2$

Combine like terms. Let's group the $m^2$ terms, the $m$ terms, and the constant terms:

$\Delta = (64m^2 - 180m^2) + 224m + (196 - 180)$

$\Delta = -116m^2 + 224m + 16$

We need this discriminant to be greater than zero for two distinct intersection points:

$-116m^2 + 224m + 16 > 0$

This is a quadratic inequality in terms of 'm'. To make it easier to work with, we can divide the entire inequality by a common factor. Let's see if we can divide by -4 (since the leading coefficient is negative, dividing by a negative number will flip the inequality sign, which is often helpful).

Dividing by -4:

$(-116m^2 / -4) + (224m / -4) + (16 / -4) < 0$

$29m^2 - 56m - 4 < 0$

So, now we need to solve the quadratic inequality $29m^2 - 56m - 4 < 0$. The first step is to find the roots of the corresponding quadratic equation $29m^2 - 56m - 4 = 0$. We can use the quadratic formula for this:

$m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 29$, $b = -56$, and $c = -4$.

Let's calculate the discriminant for this quadratic equation first (let's call it $\Delta_m$ to avoid confusion with the previous $\Delta$):

$\Delta_m = (-56)^2 - 4(29)(-4)$

$\Delta_m = 3136 - (-464)$

$\Delta_m = 3136 + 464$

$\Delta_m = 3600$

This is a perfect square! $\sqrt{3600} = 60$. That's great, it means our roots will be rational numbers.

Now, let's find the roots for 'm':

$m = \frac{-(-56) \pm \sqrt{3600}}{2(29)}$

$m = \frac{56 \pm 60}{58}$

We have two roots:

$m_1 = \frac{56 + 60}{58} = \frac{116}{58} = 2$

$m_2 = \frac{56 - 60}{58} = \frac{-4}{58} = -\frac{2}{29}$

So, the roots of $29m^2 - 56m - 4 = 0$ are $m = 2$ and $m = -2/29$.

Now, we need to solve the inequality $29m^2 - 56m - 4 < 0$. This is a parabola that opens upwards (because the coefficient of $m^2$, which is 29, is positive). A parabola that opens upwards is negative (below the x-axis) between its roots.

Therefore, the inequality $29m^2 - 56m - 4 < 0$ holds true for values of 'm' that are strictly between the two roots we found.

So, the set of values for 'm' is: $-2/29 < m < 2$.

The Final Answer and Why It Makes Sense

We have successfully found the range of 'm' values for which the line $y = mx + 1$ intersects the circle $(x - 7)^2 + (y - 5)^2 = 20$ at two distinct points. The answer is the open interval $(-2/29, 2)$. This means any slope 'm' between $-2/29$ (approximately -0.069) and $2$, exclusive, will result in our line cutting through the circle twice.

Let's quickly recap the journey. We started by substituting the line equation into the circle equation, which transformed the problem into finding the conditions for a quadratic equation in $x$ to have two distinct real roots. This condition led us to analyze a discriminant, which in turn resulted in a quadratic inequality in $m$. Solving this inequality gave us the final range for $m$. It's a step-by-step process that builds upon fundamental algebraic and geometric concepts.

Consider the endpoints: If $m = 2$ or $m = -2/29$, the discriminant is zero, meaning the line would be tangent to the circle, intersecting at exactly one point. If $m$ is outside this range (e.g., $m = 3$ or $m = -1$), the discriminant would be negative, and the line would miss the circle entirely.

This problem is a fantastic example of how seemingly complex geometric relationships can be systematically solved using algebraic tools. It highlights the power of substitution and the discriminant in determining the nature of intersections between curves. Keep practicing these types of problems, guys, and you'll become geometry and algebra masters in no time! Happy solving!